[LeetCode]268. 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?

思路：在没有元素缺失的情况下，数组下标值+1之和与数组元素之和的差为n，当有元素x缺失时，这个差变为x，因此只需要计算数组下标值+1的和与数组元素之和的差值即可

public class Solution {
    public int missingNumber(int[] nums) {
        int result = 0;
        for(int i=0; i<nums.length; i++){
            result = result + i+1 - nums[i];
        }
        return result;
    }

}