LeetCode Single Number II

鏈接: https://oj.leetcode.com/problems/single-number-ii/

雖然知道這個題肯定跟位運算,異或有關...但實在沒想出來怎麼解決

class Solution {
public:
    int singleNumber(int A[], int n) {
        int l=0,r=1;
        sort(A,A+n);
        if(n==1)
            return A[0];
        while(A[l]==A[r])
        {
            l+=3;
            r+=3;
            if(r>=n)
                break;
        }
        return A[l];
    }
};

後來google了一下

:如果別的數字都出現兩次,那麼解決方法是分別對n個數字異或(見 Single Number)..而異或的實質就是按位模2加,所以,這題的思路就是模擬按位模3加:把出現3次1的位置0,而出現1次1的置1;

該代碼來自網絡:

class Solution {
public:
    int singleNumber(int A[], int n) {
        int one = 0;
        int two = 0;
        int three = 0;
        for(int i = 0 ; i < n ; i++){
            two |= A[i] & one;
            one = A[i] ^ one;
            three = ~(one&two);
            one &= three;
            two &= three;
        }
        return one;
    }
};