In one move, you can either:

• Increment the current integer by one (i.e., x = x + 1).
• Double the current integer (i.e., x = 2 * x).

You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times.

Given the two integers target and maxDoubles, return the minimum number of moves needed to reach target starting with 1.

Example 1:

Input: target = 5, maxDoubles = 0
Output: 4
Explanation: Keep incrementing by 1 until you reach target.

Example 2:

Input: target = 19, maxDoubles = 2
Output: 7
Explanation: Initially, x = 1
Increment 3 times so x = 4
Double once so x = 8
Increment once so x = 9
Double again so x = 18
Increment once so x = 19

Example 3:

Input: target = 10, maxDoubles = 4
Output: 4
Explanation:Initially, x = 1
Increment once so x = 2
Double once so x = 4
Increment once so x = 5
Double again so x = 10

Constraints:

• 1 <= target <= 109
• 0 <= maxDoubles <= 100

Solution: Reverse + Greedy

If num is odd, decrement it by 1. Divide num by 2 until maxdoubles times. Apply decrementing until 1 reached.

ex1: 19 (dec)-> 18 (div1)-> 9 (dec) -> 8 (div2)-> 4 (dec)-> 3 (dec)-> 2 (dec)-> 1

Time complexity: O(logn)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int minMoves(int target, int maxDoubles) {
    int ans = 0;
    while (maxDoubles-- && target != 1) {
      if (target & 1)
        --target, ++ans;
      ++ans;
      target >>= 1;      
    }
    ans += (target - 1);
    return ans;
  }
};

The post 花花酱 LeetCode 2139. Minimum Moves to Reach Target Score appeared first on Huahua's Tech Road.

• For example, if n = 4, then node 0 is the twin of node 3, and node 1 is the twin of node 2. These are the only nodes with twins for n = 4.

The twin sum is defined as the sum of a node and its twin.

Given the head of a linked list with even length, return the maximum twin sum of the linked list.

Example 1:

Input: head = [5,4,2,1]
Output: 6
Explanation:
Nodes 0 and 1 are the twins of nodes 3 and 2, respectively. All have twin sum = 6.
There are no other nodes with twins in the linked list.
Thus, the maximum twin sum of the linked list is 6. 

Example 2:

Input: head = [4,2,2,3]
Output: 7
Explanation:
The nodes with twins present in this linked list are:
- Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7.
- Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4.
Thus, the maximum twin sum of the linked list is max(7, 4) = 7. 

Example 3:

Input: head = [1,100000]
Output: 100001
Explanation:
There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.

Constraints:

• The number of nodes in the list is an even integer in the range [2, 105].
• 1 <= Node.val <= 105

Solution: Two Pointers + Reverse List

Use fast slow pointers to find the middle point and reverse the second half.

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int pairSum(ListNode* head) {
    auto reverse = [](ListNode* head, ListNode* prev = nullptr) {
      while (head) {
        swap(head->next, prev);
        swap(head, prev);
      }
      return prev;
    };
    
    ListNode* fast = head;
    ListNode* slow = head;
    while (fast) {
      fast = fast->next->next;
      slow = slow->next;
    }
    
    slow = reverse(slow);
    int ans = 0;
    while (slow) {
      ans = max(ans, head->val + slow->val);
      head = head->next;
      slow = slow->next;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2130. Maximum Twin Sum of a Linked List appeared first on Huahua's Tech Road.

L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list’s nodes. Only nodes themselves may be changed.

Example 1:

Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]

Constraints:

• The number of nodes in the list is in the range [1, 5 * 104].
• 1 <= Node.val <= 1000

Solution: Three steps

Step 1: Find mid node that splits the list into two halves.
Step 2: Reverse the second half
Step 3: Merge two lists

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  void reorderList(ListNode* head) {
    if (!head || !head->next) return;
    ListNode* slow = head;
    ListNode* fast = head;            

    while (fast->next && fast->next->next) {
      slow = slow->next;
      fast = fast->next->next;          
    }
    
    ListNode* h1 = head;
    ListNode* h2 = reverse(slow->next);
    slow->next = nullptr;

    while (h1 && h2) {
      ListNode* n1 = h1->next;
      ListNode* n2 = h2->next;
      h1->next = h2;
      h2->next = n1;
      h1 = n1;
      h2 = n2;
    }
  }
private:       
  // reverse a linked list
  // returns head of the linked list
  ListNode* reverse(ListNode* head) {
    if (!head || !head->next) return head;

    ListNode* prev = head;
    ListNode* cur = head->next;
    ListNode* next = cur;
    while (cur) {
      next = next->next;
      cur->next = prev;
      prev = cur;
      cur = next;
    }

    head->next = nullptr;
    return prev;
  }
};

The post 花花酱 LeetCode 143. Reorder List appeared first on Huahua's Tech Road.

A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.

Return a string of the words in reverse order concatenated by a single space.

Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.

Example 1:

Input: s = "the sky is blue"
Output: "blue is sky the"

Example 2:

Input: s = "  hello world  "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.

Example 3:

Input: s = "a good   example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.

Example 4:

Input: s = "  Bob    Loves  Alice   "
Output: "Alice Loves Bob"

Example 5:

Input: s = "Alice does not even like bob"
Output: "bob like even not does Alice"

Constraints:

• 1 <= s.length <= 104
• s contains English letters (upper-case and lower-case), digits, and spaces ' '.
• There is at least one word in s.

Follow-up: If the string data type is mutable in your language, can you solve it in-place with O(1) extra space?

Solution: Stack

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  string reverseWords(string s) {
    stringstream ss(s);
    stack<string> st;
    string w;
    while (ss >> w)
      st.push(w);
    string ans;
    while (!st.empty()) {
      ans += st.top();
      st.pop();
      if (!st.empty()) ans += ' ';
    }
    return ans;
  }
};

Solution: In-Place

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  string reverseWords(string& s) {
    reverse(begin(s), end(s));
    int l = 0;
    for (int i = 0, j = 0; i < s.length(); ++i) {
      if (s[i] == ' ') continue;
      if (l) ++l;
      j = i;
      while (j < s.length() && s[j] != ' ') ++j;
      reverse(begin(s) + l, begin(s) + j);
      l += (j - i);
      i = j;
    }
    s.resize(l);
    return s;
  }
};

The post 花花酱 LeetCode 151. Reverse Words in a String appeared first on Huahua's Tech Road.

You are allowed to select any subarray of the given array and reverse it. You can perform this operation only once.

Find maximum possible value of the final array.

Example 1:

Input: nums = [2,3,1,5,4]
Output: 10
Explanation: By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.

Example 2:

Input: nums = [2,4,9,24,2,1,10]
Output: 68

Constraints:

• 1 <= nums.length <= 3*10^4
• -10^5 <= nums[i] <= 10^5

Solution: Greedy

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int maxValueAfterReverse(vector<int>& nums) {
    const int n = nums.size();
    int sum = 0;
    int gain = 0;    
    int hi = INT_MIN;
    int lo = INT_MAX;
    for (int i = 0; i < n - 1; ++i) {
      int n1 = nums[i];
      int n2 = nums[i + 1];
      sum += abs(n1 - n2);
      gain = max({gain, 
                 abs(nums[0] - n2) - abs(n1 - n2),
                 abs(nums[n - 1] - n1) - abs(n1 - n2)});
      hi = max(hi, min(n1, n2));
      lo = min(lo, max(n1, n2));
    }
    return sum + max(gain, (hi - lo) * 2);
  }
};

The post 花花酱 LeetCode 1330. Reverse Subarray To Maximize Array Value appeared first on Huahua's Tech Road.

Reverse bits of a given 32 bits unsigned integer.

Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.

Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 2 above the input represents the signed integer -3 and the output represents the signed integer -1073741825.

Follow up:

If this function is called many times, how would you optimize it?

Solution: Bit operation

Time complexity: O(1)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  uint32_t reverseBits(uint32_t n) {
    uint32_t ans = 0;
    for (int i = 0; i < 32; ++i) {
      ans <<= 1;
      ans |= n & 1;      
      n >>= 1;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 190. Reverse Bits appeared first on Huahua's Tech Road.

Reverse the strings in each pair of matching parentheses, starting from the innermost one.

Your result should not contain any bracket.

Example 1:

Input: s = "(abcd)"
Output: "dcba"

Example 2:

Input: s = "(u(love)i)"
Output: "iloveu"

Example 3:

Input: s = "(ed(et(oc))el)"
Output: "leetcode"

Example 4:

Input: s = "a(bcdefghijkl(mno)p)q"
Output: "apmnolkjihgfedcbq"

Constraints:

• 0 <= s.length <= 2000
• s only contains lower case English characters and parentheses.
• It’s guaranteed that all parentheses are balanced.

Solution: Stack

Use a stack of strings to track all the active strings.
Iterate over the input string:
1. Whenever there is a ‘(‘, push an empty string to the stack.
2. Whenever this is a ‘)’, pop the top string and append the reverse of it to the new stack top.
3. Otherwise, append the letter to the string on top the of stack.

Once done, the (only) string on the top of the stack is the answer.

Time complexity: O(n^2)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  string reverseParentheses(string s) {
    stack<string> st;
    st.push("");
    for (char c : s) {
      if (c == '(') st.push("");
      else if (c != ')') st.top() += c;
      else {
        string t = st.top(); st.pop();
        st.top().insert(end(st.top()), rbegin(t), rend(t));
      }
    }
    return st.top();
  }
};

The post 花花酱 LeetCode 1190. Reverse Substrings Between Each Pair of Parentheses appeared first on Huahua's Tech Road.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

Solution:

1. Store the m – 1 and m-th item as prev and tail before reversing
2. Reverse the m to n, return the head and tail->next of the reversed list
3. Reconnect prev and head, tail and tail->next

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  ListNode* reverseBetween(ListNode* head, int m, int n) {
    ListNode dummy(0);
    dummy.next = head;
    ListNode* p = &dummy;
    // Find the m-1 th node
    for (int i = 0; i < m - 1; ++i) 
      p = p->next;
    ListNode* prev = p;
    ListNode* curr = p->next;
    ListNode* tail = curr;    
    // Reverse from m to n
    for (int i = m; i <= n; ++i) {
      ListNode* next = curr->next;
      curr->next = prev;
      prev = curr;
      curr = next;
    }    
    //  Connect three parts
    p->next = prev;
    tail->next = curr;
    return dummy.next;
  }
};

# Author: Huahua
class Solution:
  def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
    dummy = ListNode(0)
    dummy.next = head
    p = dummy
    for i in range(m - 1):
      p = p.next
    prev = p
    curr = p.next
    tail = curr
    for i in range(n - m + 1):
      next = curr.next
      curr.next = prev
      prev = curr
      curr = next
    p.next = prev
    tail.next = curr
    return dummy.next

The post 花花酱 LeetCode 92. Reverse Linked List II appeared first on Huahua's Tech Road.

Given a string S, return the “reversed” string where all characters that are not a letter stay in the same place, and all letters reverse their positions.

Example 1:

Input: "ab-cd"
Output: "dc-ba"

Example 2:

Input: "a-bC-dEf-ghIj"
Output: "j-Ih-gfE-dCba"

Example 3:

Input: "Test1ng-Leet=code-Q!"
Output: "Qedo1ct-eeLg=ntse-T!"

Note:

1. S.length <= 100
2. 33 <= S[i].ASCIIcode <= 122
3. S doesn’t contain \ or "

Solution: Two Pointers

Time complexity: O(n)

Space complexity: O(1) – in place

// Author: Huahua
class Solution {
public:
  string reverseOnlyLetters(string S) {
    int i = 0;
    int j = S.length() - 1;
    while (i < j) {
      if (isalpha(S[i]) && isalpha(S[j])) {
        swap(S[i++], S[j--]);        
      } else {
        if (!isalpha(S[i])) ++i;
        if (!isalpha(S[j])) --j;
      }
    }
    return S;
  }
};

// Author: Huahua
class Solution {
public:
  string reverseOnlyLetters(string S) {
    auto it1 = begin(S);
    auto it2 = prev(end(S));
    while (it1 < it2) {
      if (isalpha(*it1) && isalpha(*it2)) {
        swap(*it1++, *it2--);
      } else {
        if (!isalpha(*it1)) ++it1;
        if (!isalpha(*it2)) --it2;
      }
    }
    return S;
  }
};

The post 花花酱 LeetCode 917. Reverse Only Letters appeared first on Huahua's Tech Road.

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place and use only constant extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.

1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

Solution

Find the last acceding element x, swap with the smallest number y, y is after x that and y is greater than x.

Reverse the elements after x.

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua, 8 ms
class Solution {
public:
  void nextPermutation(vector<int>& nums) {
    int i = nums.size() - 2;
    while (i >= 0 && nums[i + 1] <= nums[i]) --i;
    if (i >= 0) {
      int j = nums.size() - 1;
      while (j >= 0 && nums[j] <= nums[i]) --j;
      swap(nums[i], nums[j]);
    }
    reverse(begin(nums) + i + 1, end(nums));
  }
};

# Author: Huahua, 48 ms
class Solution:
  def nextPermutation(self, nums):
    n = len(nums)
    i = n - 2
    while i >= 0 and nums[i] >= nums[i + 1]: i -= 1
    if i >= 0:
      j = n - 1
      while j >= 0 and nums[j] <= nums[i]: j -= 1
      nums[i], nums[j] = nums[j], nums[i]
    nums[i + 1:] = nums[i+1:][::-1]

The post 花花酱 LeetCode 31. Next Permutation appeared first on Huahua's Tech Road.

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹,  2³¹ − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution: Simulation

Reverse digit by digit. Be careful about the overflow and negative numbers (especially in Python)

Time complexity: O(log(x)) ~ O(1)

Space complexity: O(log(x)) ~ O(1)

// Author: Huahua
class Solution {
public:
  int reverse(int x) {
    int ans = 0;
    while (x != 0) {
      int r = x % 10;
      if (ans > INT_MAX / 10 || ans < INT_MIN / 10) return 0;
      ans = ans * 10 + r;
      x /= 10;
    }
    return ans;
  }
};

class Solution {
  public int reverse(int x) {
    int ans = 0;
    while (x != 0) {
      int r = x % 10;
      if (ans > Integer.MAX_VALUE / 10 || ans < Integer.MIN_VALUE / 10) return 0;
      ans = ans * 10 + r;
      x /= 10;
    }
    return ans;
  }
}

class Solution:
  def reverse(self, x):
    min_val = -(2**31);
    max_val = 2**31 - 1;
    sign = -1 if x < 0 else 1
    x = abs(x)
    ans = 0
    while x != 0:            
      ans = ans * 10 + (x % 10)
      x //= 10
    ans *= sign
    if ans > max_val or ans < min_val: return 0
    return ans

The post 花花酱 LeetCode 7. Reverse Integer appeared first on Huahua's Tech Road.

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

Input: [1,2,3,4,5,6,7] and k = 3 
Output: [5,6,7,1,2,3,4]
Explanation: rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4,5] 
rotate 3 steps to the right: [5,6,7,1,2,3,4]

Example 2:

Input:[-1,-100,3,99] and k = 2 
Output: [3,99,-1,-100] 
Explanation: rotate 1 steps to the right: [99,-1,-100,3] rotate 2 steps to the right: [3,99,-1,-100]

Note:

• Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
• Could you do it in-place with O(1) extra space?

Solution 1: Simulate rotation with three reverses.

If k >= n, rotating k times has the same effect as rotating k % n times.

[1,2,3,4,5,6,7], K = 3

[5,6,7,1,2,3,4]

We can simulate the rotation with three reverses.

1. reverse the whole array O(n) [7,6,5,4,3,2,1]
2. reverse the left part 0 ~ k – 1 O(k) [5,6,7,4,3,2,1]
3. reverse the right part k ~ n – 1 O(n-k) [5,6,7,1,2,3,4]

Time complexity: O(n)

Space complexity: O(1) in-place

C++

class Solution {
public:
  void rotate(vector<int>& nums, int k) {
    if (nums.empty()) return;
    k %= nums.size();
    if (k == 0) return;
    reverse(begin(nums), end(nums));
    reverse(begin(nums), begin(nums) + k);
    reverse(begin(nums) + k, end(nums));
  }
};

The post 花花酱 LeetCode 189. Rotate Array appeared first on Huahua's Tech Road.

Write a function that takes a string as input and reverse only the vowels of a string.

Example 1:
Given s = “hello”, return “holle”.

Example 2:
Given s = “leetcode”, return “leotcede”.

Note:
The vowels does not include the letter “y”.

Solution

// Author: Huahua
// Running time: 8 ms
class Solution {
public:
    string reverseVowels(string s) {
      int i = 0;
      int j = s.size() - 1;
      while (i < j) {
        while (!isVowel(s[i]) && i < j) ++i;
        while (!isVowel(s[j]) && i < j) --j;                
        swap(s[i++], s[j--]);
      }
      return s;
    }
private:
  bool isVowel(char c) {
    c = tolower(c);
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
  }
};

The post 花花酱 LeetCode 345. Reverse Vowels of a String appeared first on Huahua's Tech Road.

Problem

题目大意：把字符串分成块，每块长度2k，单独反转每一块的前k的字符。

https://leetcode.com/problems/reverse-string-ii/description/

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

Solution

C++

// Author: Huahua
// Running time: 9 ms
class Solution {
public:
  string reverseStr(string s, int k) {
    const int n = s.length();
    for (int i = 0; i <= n / k; i += 2)      
      std::reverse(s.begin() + i * k, s.begin() + min(n, (i + 1) * k));
    return s;
  }
};

Related Problems

• 花花酱 LeetCode 557 Reverse Words in a String III

The post 花花酱 LeetCode 541. Reverse String II appeared first on Huahua's Tech Road.

Problem:

https://leetcode.com/problems/reverse-linked-list/description/

Reverse a singly linked list.

Solution 1:

Tracking prev / curr / next node

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua
// Running time: 9 ms
class Solution {
public:
  ListNode* reverseList(ListNode* head) {
    ListNode* prev = nullptr;
    ListNode* curr = head;
    ListNode* next;
    while (curr) {
      next = curr->next;
      curr->next = prev;
      prev = curr;
      curr = next;
    }
    return prev;
  }
};

Java

// Author: Huahua
// Running time: 0 ms
class Solution {
  public ListNode reverseList(ListNode head) {
    ListNode prev = null;
    ListNode curr = head;
    ListNode next;
    while (curr != null) {
      next = curr.next;
      curr.next = prev;
      prev = curr;
      curr = next;
    }
    return prev;
  }
}

Python 3

"""
Author: Huahua
Running time: 48 ms
"""
class Solution:
  def reverseList(self, head):
    prev, curr, nxt = None, head, None;    
    while curr:
      nxt, curr.next = curr.next, prev      
      prev, curr = curr, nxt
    return prev

The post 花花酱 LeetCode 206. Reverse Linked List appeared first on Huahua's Tech Road.