leetcode - Single Number II

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leetcode - 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?

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

用一个32维的数组，存储每一位上1出现的次数，最后将每一位上的数取3的余数，为所求数字在该为上的数字，

例如第5位上得到21，除3取余数得0，说明，所求的数字在第5位为0

最后恢复出所求数字即可。

时间复杂度：O(32*N)

http://blog.csdn.net/jiadebin890724/article/details/23306837

leetcode - Single Number II

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原文地址：http://www.cnblogs.com/shnj/p/4739752.html