题目链接：

题目：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?

解题思路：题意为给定一个包括n个不反复的数的数组。从0,1,2...n，找出数组中遗漏的那个数。

演示样例代码例如以下：

public class Solution{	public int missingNumber(int[] nums) 	{		//首先对数组进行排序		Arrays.sort(nums);		int startData=nums[0];		for(int i=1;i
     
      0)				return 0;			else				return nums[nums.length-1]+1;		}		return 0;    }}