花花酱 LeetCode 7. Reverse Integer

By zxi on September 12, 2018

Problem

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)

C++

// 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;

}

};

Java

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;

}

Python3

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

花花酱 LeetCode 6. ZigZag Conversion

By zxi on September 12, 2018

Problem

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:
<preclass=”crayon:false”>Input: s = “PAYPALISHIRING”, numRows = 3 Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Solution: Simulation

Store the zigzag results for each row and then output row by row.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

string convert(string s, int nRows) {

if (nRows == 1) return s;

vector<string> ss(nRows);

int l = s.length();

int x = 0, y = 0, i = 0;

bool down = true;

while (i < l) {

ss[y] += s[i];

if (down) {

++y;

if (y == nRows) {

down = false;

y -=2;

}

} else {

--y;

if (y < 0) {

down = true;

y = 1;

}

i++;

}

string ans;

for(const string& r : ss)

ans += r;

return ans;

}

};

花花酱 LeetCode 5. Longest Palindromic Substring

By zxi on September 12, 2018

Problem

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

Solution: DP

Try all possible i and find the longest palindromic string whose center is i (odd case) and i / i + 1 (even case).

Time complexity: O(n^2)

Space complexity: O(1)

C++

// Author: Huahua, 16 ms, 8.7 MB

class Solution {

public:

string longestPalindrome(string s) {

const int n = s.length();

auto getLen = [&](int l, int r) {

while (l >= 0 && r < n

&& s[l] == s[r]) {

--l;

++r;

}

return r - l - 1;

};

int len = 0;

int start = 0;

for (int i = 0; i < n; ++i) {

int cur = max(getLen(i, i),

getLen(i, i + 1));

if (cur > len) {

len = cur;

start = i - (len - 1) / 2;

}

return s.substr(start, len);

}

};

Java

// Author: Huahua, 6 ms， 36.5 MB

class Solution {

public String longestPalindrome(String s) {

int len = 0;

int start = 0;

for (int i = 0; i < s.length(); ++i) {

int cur = Math.max(getLen(s, i, i),

getLen(s, i, i + 1));

if (cur > len) {

len = cur;

start = i - (cur - 1) / 2;

}

return s.substring(start, start + len);

}

private int getLen(String s, int l, int r) {

while (l >= 0 && r < s.length()

&& s.charAt(l) == s.charAt(r)) {

--l;

++r;

}

return r - l - 1;

}

Python3

# Author: Huahua, 812 ms, 12.7MB

class Solution:

def longestPalindrome(self, s: str) -> str:

n = len(s)

def getLen(l, r):

while l >= 0 and r < n and s[l] == s[r]:

l -= 1

r += 1

return r - l - 1

start = 0

length = 0

for i in range(n):

cur = max(getLen(i, i),

getLen(i, i + 1))

if cur <= length: continue

length = cur

start = i - (cur - 1) // 2

return s[start : start + length]

花花酱 LeetCode 3. Longest Substring Without Repeating Characters

By zxi on September 11, 2018

Problem

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3.

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

Solution: HashTable + Sliding Window

Using a hashtable to remember the last index of every char. And keep tracking the starting point of a valid substring.

start = max(start, last[s[i]] + 1)

ans = max(ans, i – start + 1)

Time complexity: O(n)

Space complexity: O(128)

C++

// Author: Huahua

class Solution {

public:

int lengthOfLongestSubstring(string s) {

const int n = s.length();

int ans = 0;

vector<int> idx(128, -1);

for (int i = 0, j = 0; j < n; ++j) {

i = max(i,idx[s[j]] + 1);

ans = max(ans, j - i + 1);

idx[s[j]] = j;

}

return ans;

}

};

Java

// Author: Huahua

class Solution {

public int lengthOfLongestSubstring(String s) {

int[] last = new int[128];

Arrays.fill(last, -1);

int start = 0;

int ans = 0;

for (int i = 0; i < s.length(); ++i) {

if (last[s.charAt(i)] != -1)

start = Math.max(start, last[s.charAt(i)] + 1);

last[s.charAt(i)] = i;

ans = Math.max(ans, i - start + 1);

}

return ans;

}

Python3

# Author: Huahua

class Solution:

def lengthOfLongestSubstring(self, s):

last = [-1] * 128

ans = 0

start = 0

for i, ch in enumerate(s):

if last[ord(ch)] != -1:

start = max(start, last[ord(ch)] + 1)

ans = max(ans, i - start + 1)

last[ord(ch)] = i

return ans

花花酱 LeetCode 668. Kth Smallest Number in Multiplication Table

By zxi on September 10, 2018

Problem

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table?

Given the height m and the length n of a m * n Multiplication Table, and a positive integer k, you need to return the k-th smallest number in this table.

Example 1:

Input: m = 3, n = 3, k = 5
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6
3	6	9

The 5-th smallest number is 3 (1, 2, 2, 3, 3).

Example 2:

Input: m = 2, n = 3, k = 6
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6

The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).

Note:

The m and n will be in the range [1, 30000].
The k will be in the range [1, m * n]

Solution 1: Nth element (MLE)

Time complexity: O(mn)

Space complexity: O(mn)

C++

// 59 / 69 test cases passed.

class Solution {

public:

int findKthNumber(int m, int n, int k) {

vector<int> s(m * n);

auto it = begin(s);

for (int i = 1; i <= m; ++i)

for (int j = 1; j <= n; ++j)

*it++ = i * j;

nth_element(begin(s), begin(s) + k - 1, end(s));

return s[k - 1];

}

};

Solution2 : Binary Search

Find first x such that there are k elements less or equal to x in the table.

Time complexity: O(m*log(m*n))

Space complexity: O(1)

C++

// Author: Huahua, 12 ms

class Solution {

public:

int findKthNumber(int m, int n, int k) {

int l = 1;

int r = m * n + 1;

while (l < r) {

int x = l + (r - l) / 2;

if (LEX(m, n, x) >= k)

r = x;

else

l = x + 1;

}

return l;

}

private:

// Returns # of elements in the m*n table that are <= x

int LEX(int m, int n, int x) {

int count = 0;

for (int i = 1; i <= m; ++i)

count += min(n, x / i);

return count;

}

};

Java

// Author: Huahua, 16 ms

class Solution {

public int findKthNumber(int m, int n, int k) {

int l = 1;

int r = m * n + 1;

while (l < r) {

int x = l + (r - l) / 2;

if (LEX(m, n, x) >= k)

r = x;

else

l = x + 1;

}

return l;

}

// Returns # of elements in the m*n table that are <= x

private int LEX(int m, int n, int x) {

int count = 0;

for (int i = 1; i <= m; ++i)

count += Math.min(n, x / i);

return count;

}

Python3

# Author: Huahua, 976 ms

class Solution:

def findKthNumber(self, m, n, k):

def LEX(m, n, x, k):

count = 0

for i in range(1, min(m + 1, x + 1)):

count += min(n, x // i)

if count >= k: return count

return count

l = 1

r = m * n + 1

while l < r:

x = l + (r - l) // 2

if LEX(m, n, x, k) >= k:

r = x

else:

l = x + 1

return l

Huahua's Tech Road

花花酱 LeetCode 7. Reverse Integer

Problem

Solution: Simulation

C++

Java

Python3

花花酱 LeetCode 6. ZigZag Conversion

Problem

Solution: Simulation

C++

花花酱 LeetCode 5. Longest Palindromic Substring

Problem

Solution: DP

C++

Java

Python3

花花酱 LeetCode 3. Longest Substring Without Repeating Characters

Problem

Solution: HashTable + Sliding Window

C++

Java

Python3

花花酱 LeetCode 668. Kth Smallest Number in Multiplication Table

Problem

Solution 1: Nth element (MLE)

C++

Solution2 : Binary Search

C++

Java

Python3

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