[LeetCode] Single Number II

Single Number II

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?

解题思路：

与Single Number不同，这里不能再用异或操作来做了。可以为每个位上1的数目计数。然后将该计数%3，剩下的1肯定是单独的数字贡献的。

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        vector<int> bitNumCount(32, 0);
        int len = nums.size();
        for(int i=0; i<len; i++){
            int num = nums[i];
            for(int j = 0; j<32 && num!=0; j++){
                if(num & 0x1 == 1){
                    bitNumCount[j]++;
                }
                num = num>>1;
            }
        }
        int result = 0;
        for(int i=31; i>=0; i--){
            result = result<<1;
            if(bitNumCount[i]%3!=0){
                result += 1;
            }
        }
        return result;
    }
};

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int result = 0;
        int len = nums.size();
        for(int i=0; i<32; i++){
            int count = 0;
            for(int j=0; j<len; j++){
                count += (nums[j]>>i)&1;
            }
            result += (count%3)<<i;
        }
        
        return result;
    }
};

[LeetCode] Single Number II

原文：http://blog.csdn.net/kangrydotnet/article/details/46560863