[leetcode] Missing Number

题目：

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?

分析：

方法一：根据元素的顺序，进行异或^操作：

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;
}

方法二：利用数学求和，用期望值减去实际值即可：

    public int missingNumber(int[] nums) {
        int sum = 0;
        for(int i : nums) {
            sum += i;
        }
        int n = nums.length;
        int tmp = (n+1)*n/2;
        return tmp-sum;
    }