LeetCode题解：Missing Number

Given an array containing n distinct numbers taken from 0, 1, 2, …, n, find the one that is missing from the array.

For example,

Given nums = [0, 1, 3] return 2.

Note:

Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

题意：给定数组包含n个不同的数：0,1,2,……,n，找到数组中丢失的数

解题思路：第一反应是对数组求和，用求和公式减去该和……看了解题讨论发现更高效率的办法：由于数与数组索引是1对1匹配的，值也相同，那么用异或运算不就能得到丢失的数么。变换一下这题和Single Number相同。

代码：

public class Solution {
public int missingNumber(int[] nums) {
int xor = 0, i = 0;
for (i = 0; i < nums.length; i++) {
xor = xor ^ i ^ nums[i];
}

return xor ^ i;
}
}