题目描述

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?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

分析

既然是0~n只差其中一个数，不如再次加入0~n的序列，这样就和LeetCode 136 Single Number一样了，missing number只出现一次，其余都是出现两次。

代码

    public int missingNumber(int[] nums) {

        int rt = nums.length;

        for (int i = 0; i < nums.length; i++) {
            rt = rt ^ nums[i] ^ i;
        }

        return rt;
    }