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

1. 缺少的数 == sum(0.....n) - sum(num[i])
2. 有可能input的数组很大，直接计算sum会超过integer的范围，所以一边加i一边减去nums[i]。最后剩下的数就是missing number

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