268. Missing Number

问题链接：https://leetcode.com/problems/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?
刚开始采用新建一个标志位数组的方式，数组长度为nums.length+1，flag[nums[i]] i=0...n-1当出现那个数组元素为最初填充的-1时表示该位为missing
但是效率太低且空间复杂度为线性。

运用n个数的累加和的方式完成。时间常数，空间线性

public class Solution {
public int missingNumber(int[] nums) {
if(nums==null || nums.length==0)
return 0;

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