LeetCode 268 Missing Number

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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?

给定一个含有n个数的数组，元素的范围是0~n，数组长度为n，而不是n+1，所以数组丢失了一个元素，请找到丢失的一个元素。数组不一定是排序的。

第一种做法：使用位运算，注意最后还要和n进行处理，因为数组的最后一位坐标为n-1.

public int missingNumber(int[] nums) {int result = 0;for (int i = 0; i < nums.length; i++) {result = result^nums[i];}return result^nums.length;}

第二种做法：

public int missingNumber2(int[] nums) {int result = 0;for (int i = 0; i < nums.length; i++) {result += i - nums[i];}return result + nums.length;}

第三种做法:

public int missingNumber2(int[] nums) {int result = (1 + nums.length) * nums.length / 2;for (int i = 0; i < nums.length; i++) {result -= nums[i];}return result;}

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