LeetCode-268. Missing Number (Java)

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?

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思路

按照我的思路，先对数组进行了排序，然后对nums长度进行判断，然后对nums的元素进行判断，然后循环。。。。。。

最后耗时惨不忍睹。最后参考一下其他已提交的代码。

代码

public class Solution {
public int missingNumber(int[] nums) {
int sum = 0;
int n = nums.length ;
int i = 0;
while(i < nums.length)
{
sum += nums[i];
i++;
}
return (1+n)*n/2 - sum;
}
}

这个对数组进行求和，然后与等差公式计算的和求差，计算结果即missing number。