Given a string s, return the longest palindromic substring in s.

Constraints:

1 <= s.length <= 1000

s consist of only digits and English letters.

v1.0

class Solution {
public:
    void check(string s, int left, int right, int& Maxlen, int& start){
        while(left>=0 && right<s.length() && s[left] == s[right]){
            left--;
            right++;
        }
        if(Maxlen < right-left+1){
            start = left + 1 ;
            Maxlen = right-left+1;
        }
    }
    
    string longestPalindrome(string s) {
        int Maxlen=0;
        int start=0;
        
        for(int i=0 ; i<s.length() ; i++){
            check(s,i,i,Maxlen,start);
            check(s,i,i+1,Maxlen,start);
        }
        return s.substr(start, Maxlen-2);
    }
};

參考網路寫法。

以每個字元都做為中心，漸漸往左右發散。
要分兩個狀況：1. odd palindrome 2. even palindrome

小瑕疵：計算出的左邊界與右邊界不會是正確的，會再往左/右多出一個，因此最大長度要再減2。

Runtime: 264 ms, faster than 48.39% of C++ online submissions for Longest Palindromic Substring.

Memory Usage: 231.6 MB, less than 17.80% of C++ online submissions for Longest Palindromic Substring.

The complement of an integer is the integer you get when you flip all the 0's to 1's and all the 1's to 0's in its binary representation.

For example, The integer 5 is "101" in binary and its complement is "010" which is the integer 2.

Given an integer num, return its complement.

Constraints:

1 <= num < 2^31

v1.0

class Solution {
public:
    int findComplement(int num) {
        bitset<32> num2 = num;
        num2 = ~num2;
        bool flag = false;
        for(int i=31 ; i>=0 ; i--){
            if(num2[i]==0){
                flag = true;
            }
            else if(num2[i]==1 & flag==false){
                num2[i]=0;
            }
        }
        num = num2.to_ulong();
        return num;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Number Complement.

Memory Usage: 5.9 MB, less than 37.54% of C++ online submissions for Number Complement.

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, return the Hamming distance between them.

Constraints:

0 <= x, y <= 2^31 - 1

v1.0

class Solution {
public:
    int hammingDistance(int x, int y) {
        bitset <32> xbit = x;
        bitset <32> ybit = y;
        xbit = xbit xor ybit;
        return xbit.count();
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Hamming Distance.

Memory Usage: 5.9 MB, less than 74.11% of C++ online submissions for Hamming Distance.