Posts tagged as “easy”

花花酱 LeetCode 88. Merge Sorted Array

By zxi on September 27, 2018

Problem

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.

Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

Solution:

Fill nums1 from back to front

Time complexity: O(m + n)

Space complexity: O(1) in-place

C++

// Author: Huahua

class Solution {

public:

void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {

int i = m - 1;

int j = n - 1;

int tail = m + n - 1;

while (j >= 0)

nums1[tail--] = (i >= 0 && nums1[i] > nums2[j]) ? nums1[i--] : nums2[j--];

}

};

花花酱 LeetCode 13. Roman to Integer

By zxi on September 19, 2018

Problem

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: "III"
Output: 3

Example 2:

Input: "IV"
Output: 4

Example 3:

Input: "IX"
Output: 9

Example 4:

Input: "LVIII"
Output: 58
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: "MCMXCIV"
Output: 1994
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Solution

accumulate the value of each letter.

If the value of current letter is greater than the previous one, deduct twice of the previous value.

e.g. IX, 1 + 10 – 2 * 1 = 9 instead of 1 + 10 = 11

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int romanToInt(string s) {

map<char, int> m{{'I', 1}, {'V', 5},

{'X', 10}, {'L', 50},

{'C', 100}, {'D', 500},

{'M', 1000}};

int ans = 0;

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

ans += m[s[i]];

if (i > 0 && m[s[i]] > m[s[i - 1]])

ans -= 2 * m[s[i - 1]];

}

return ans;

}

};

Java

// Author: Huahua

class Solution {

private int v(char R) {

switch (R) {

case 'I' : return 1;

case 'V' : return 5;

case 'X' : return 10;

case 'L' : return 50;

case 'C' : return 100;

case 'D' : return 500;

case 'M' : return 1000;

default: break;

}

return 0;

}

public int romanToInt(String s) {

int ans = 0;

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

ans += v(s.charAt(i));

if ( i > 0 && v(s.charAt(i)) > v(s.charAt(i - 1)))

ans -= 2 * v(s.charAt(i - 1));

}

return ans;

}

Python3

class Solution:

def romanToInt(self, s):

m = {'I' : 1, 'V': 5, 'X': 10, 'L' : 50,

'C' : 100, 'D': 500, 'M': 1000}

ans = m[s[0]]

for c, p in zip(s[1:], s[0:-1]):

ans += m[c]

if m[c] > m[p]: ans -= 2 * m[p]

return ans

花花酱 LeetCode 905. Sort Array By Parity

By zxi on September 15, 2018

Problem

Given an array A of non-negative integers, return an array consisting of all the even elements of A, followed by all the odd elements of A.

You may return any answer array that satisfies this condition.

Example 1:

Input: [3,1,2,4]

Output: [2,4,3,1]

The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Note:

1 <= A.length <= 5000
0 <= A[i] <= 5000

Solution 1: Split Odd/Even

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua, 44 ms

class Solution {

public:

vector<int> sortArrayByParity(vector<int>& A) {

vector<int> even;

vector<int> odd;

for (int a : A) {

if (a % 2) odd.push_back(a);

else even.push_back(a);

}

even.insert(end(even), begin(odd), end(odd));

return even;

}

};

Solution 2: Stable sort by key % 2

Time complexity: O(nlogn)

Space complexity: O(1) in-place

C++

// Author: Huahua, 52 ms

class Solution {

public:

vector<int> sortArrayByParity(vector<int>& A) {

stable_sort(begin(A), end(A), [](int a, int b){

return a % 2 < b % 2;

});

return A;

}

};

花花酱 LeetCode 9. Palindrome Number

By zxi on September 12, 2018

Problem

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

Input: -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Could you solve it without converting the integer to a string?

Solution 1: Convert to string (cheating)

Time complexity: O(log10(x))

Space complexity: O(log10(x))

C++

// Author: Huahua

class Solution {

public:

bool isPalindrome(int x) {

string s = to_string(x);

return s == string(rbegin(s), rend(s));

}

};

Solution 2: Digit by Digit

Every time we compare the first and last digits of x, if they are not the same, return false. Otherwise, remove first and last digit and continue this process.

How can we achieve that via int math?

e.g. x = 9999, t = pow((10, int)log10(x)) = 1000

first digit: x / t, last digit: x % 10

then x = (x – x / t * t) / 10 removes first and last digits.

t /= 100 since we removed two digits.

x / t = 9 = 9 = x % 10, 9999 => 99

9 = 9, 99 => “”

Time complexity: O(log10(x) / 2)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool isPalindrome(int x) {

if (x < 0) return false;

int d = static_cast<int>(log10(x) + 1);

int t = pow(10, d - 1);

for (int i = 0; i < d / 2; ++i) {

if (x / t != x % 10) return false;

x = (x - x / t * t) / 10;

t /= 100;

}

return true;

}

};

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