【LeetCode OJ 268】Missing Number

【LeetCode OJ 268】Missing Number

題目：Given an array containingndistinct numbers taken from0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Givennums=[0, 1, 3]return2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

解題思路：題意為給定一個包含n個不重復的數的數組，從0,1,2...n，找出數組中遺漏的那個數。示例代碼如下：

public class Solution
{
	public int missingNumber(int[] nums) 
	{
		//首先對數組進行排序
		Arrays.sort(nums);
		int startData=nums[0];
		for(int i=1;i<nums.length;i++) startdata="=nums[nums.length-1])" else="" return="">0)
				return 0;
			else
				return nums[nums.length-1]+1;
		}
		return 0;
    }
}</nums.length;i++)>