标签：
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?
分析一：利用数学公式：1+2+3+4+...+n = n(n+1)/2
要找的数就是和减去已有的数。
class Solution { public: int missingNumber(vector<int>& nums) { int count = nums.size(); int sum = count * (count + 1) / 2; for (vector<int>::iterator iter = nums.begin(); iter != nums.end(); ++iter) { sum -= *iter; } return sum; } };
分析二：利用位操作。
public int missingNumber(int[] nums) { //xor int res = nums.length; for(int i=0; i<nums.length; i++){ res ^= i; res ^= nums[i]; } return res; }
标签：
原文地址：http://www.cnblogs.com/vincently/p/4800917.html
