[leetcode]Single Number II

时间：2014-08-07 23:00:14 收藏：0 阅读：275

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?

算法思路：

思路1：用一个数组来模拟CPU位数，记录int类型的32位中每一位出现的次数，容易理解，可惜的是没有达到题中要求的O(1)空间。

 1 public class Solution {
 2     public int singleNumber(int[] a) {
 3         int[] binary = new int[32];
 4         for(int tem : a){
 5             int bin = 1;
 6             for(int i = 0; i < 32; i++){
 7                 if((tem & bin) == bin)
 8                     binary[i]++;
 9                 bin = bin << 1;
10             }
11         }
12         int res = 0;
13         for(int i = 0; i < 32; i++){
14             res |= (binary[i] % 3) << i;
15         }
16         return res;
17     }
18 }

思路2：对思路1的优化，对每一位进行数组的遍历，求出该位出现的次数，然后%3取模，对结果的该位进行赋值。

 1 public class Solution {
 2     public int singleNumber(int[] a) {
 3         int res = 0;
 4         for(int i = 0; i < 32; i++){
 5             int binary = 1 << i;
 6             int count  = 0;
 7             for(int tem: a){
 8                 if((tem & binary) != 0){
 9                     count++;
10                 } 
11             }
12             res |= (count % 3) << i;
13         }
14         return res;
15     }
16 }

参考资料：http://www.cnblogs.com/jdflyfly/p/3828929.html

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