Given an array of integers, every element appears three times except for one. Find that single one.

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

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• 设置一个32位的数组，然然后对数组，Array【i】代表i位的1个数，统计整个数组nums【】的数字，对于每个Array【i】做mod3的运算
• 遍历nums【】数组时候，是左移&1取出，加到Array【i】上面，然后mod3
• 最后右移加到result上面

// Single Number II
// 方法1，时间复杂度O(n)，空间复杂度O(1)
public class Solution {
    public int singleNumber(int[] nums) {
        final int W = Integer.SIZE; // 一个整数的bit数，即整数字长
        int[] count = new int[W];  // count[i]表示在在i位出现的1的次数
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < W; j++) {
                count[j] += (nums[i] >> j) & 1;
                count[j] %= 3;
            }
        }
        int result = 0;
        for (int i = 0; i < W; i++) {
            result += (count[i] << i);
        }
        return result;
    }
};