Leetcode(137) Single Number II

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Description

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

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Input: [2,2,3,2]
Output: 3

Example 2:

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Input: [0,1,0,1,0,1,99]
Output: 99

解法

又是一个巧妙的位运算的题，对于每个数的二进制的每一位上的数相加，因为所有的数都出现了3次，除了1个数，那么如果这一位上的数的和不能被3整除，则那个只出现过一次的数的这一位一定是1，否则那一位是0。

具体代码如下：

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class Solution {
    public int singleNumber(int[] nums) {
        int res = 0;
        for(int i = 0; i < 32; i++) {
            int sum = 0;
            for(int j = 0; j < nums.length; j++){
                sum += (nums[j] >> i) & 1;
            }
            if(sum % 3 != 0){
                res |= 1<<i;
            }
            else{
                res |= 0<<i;
            }
        }
        return res;
    }
}