[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?

解题思路

因为java里int始终占4个字节，32位。所以我们可以对外层循环遍历32次，然后内层循环记录0-31位每一位出现的次数，内层循环结束后将结果取余于3即为当前位的值。

时间复杂度O(32 * n), 空间复杂度O(1)

代码

public int singleNumber(int[] A) {
         int bit, result = 0;  
	  
	    for (int i = 0; i < 32; i++) {  
	    	bit = 0;
	        for (int j = 0; j < A.length; j++) {  
	            if (((A[j] >> i) & 1) == 1) {  
	                bit++;  
	            }  
	        }  
	  
	        bit = bit % 3;  
	  
	        result |= bit << i;  
	    }  
	  
	    return result; 
    }