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?

public class Solution {
public int missingNumber(int[] nums) {
if(nums.length==0)
return 0;
int res=nums.length;//zhuyi!
for(int i=0;i<nums.length;i++)
{
res=res+(i-nums[i]);//i的值相加一定比nums[i]的值相加小，而且nums[i]比i新增加的元素一定是nums.length，res的初始值设为nums.length，相当于把i从0加到nums.length，而nums[i]从0到nums.length并且少了一个值，相减最后一定是所少的值
}
return res;
}
}