[LeetCode]Single Number II

题目描述:

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?

解题方案:

该题目的一般方法是开辟一个空间为32的数组bitsum[32],然后遍历题目中给定的数组,将数组中每个整数的每一bit位给分出来,加到bitsum的对应位置。最后再遍历bitsum数组,同时求bitsum[i]%3,将结果做移位操作!下面是该题的代码:

 1 class Solution {
 2 public:
 3     int singleNumber(int A[], int n) {
 4         int bitsum[32] = {0};
 5         int result = 0;
 6         for (int i = 0; i < n; ++i) {
 7             for (int j = 0; j < 32; ++j) {
 8                 bitsum[j] += A[i]>>j & 1;
 9             }
10         }
11         for (int i = 0; i < 32; ++i) {
12             result |=bitsum[i] % 3 << i;
13         }
14         return result;
15     }
16 };

 

posted @ 2014-09-27 16:26  skycore  阅读(122)  评论(0编辑  收藏  举报