LeetCode Single Number II

class Solution {
public:
    int singleNumber(int A[], int n) {
        sort(A, A+n);
        int last = A[0];
        int time = 1;
        for (int i=1; i<n; i++) {
            if (last != A[i]) {
                if(time == 3) {
                    time = 1;
                    last = A[i];
                    continue;
                }
                return last;
            }
            time++;
        }
        if (time != 3) return last;
    }
};

一时想不出线性的方法

第二轮：

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

class Solution {
public:
    int singleNumber(int A[], int n) {
        int bitcount[32] = {0};
        for (int i=0; i<n; i++) {
            int cur = A[i];
            for (int i=0; i<32 & cur != 0; i++) {
                bitcount[i] += cur & 0x1;
                cur>>=1;
            }
        }
        int num = 0;
        for (int i=32-1; i>=0; i--) {
            num<<=1;
            num|=bitcount[i] % 3 != 0;
        }
        return num;
    }
};

 还是用这个方法吧。