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?
数组是从0—N的，中间少了一个数字。单独是{0}时候少的是1，单独是{1}时候少的是0
用一个数组记录下来0~N出现与否，该数组下标 i 表示数字 i 是否出现，出现过为1。
public class Solution { public int MissingNumber(int[] nums) { int missingNumber = -1; int[] count = new int[nums.Count() + 1]; for(int i = 0; i < nums.Count(); i++) { count[nums[i]] = 1; } for(int i = 0; i < nums.Count() + 1; i++) { if(count[i] != 1) { missingNumber = i; break; } } return missingNumber; } }
原文：http://www.cnblogs.com/Anthony-Wang/p/5095861.html