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Given an unsorted integer array, find the first missing positive integer.
For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.
Your algorithm should run in O(n) time and uses constant space.
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大概意思：找出第一个缺失的正整数，
由题意缺失正整数n,只能1<= n <=nums.size()+1,所以可以通过类似hash的方式，将nums的值与下标进行映射，最终让正整数部分有序。
class Solution { public: int firstMissingPositive(vector<int>& nums) { int len=nums.size(); for(int i=0;i<len;i++){ while(nums[i]>0&&nums[i]<=len&&nums[i]!=nums[nums[i]-1]){ swap(nums[i],nums[nums[i]-1]); } } for(int i=0;i<len;i++){ if(nums[i]!=i+1){ return i+1; } } return len+1; } };
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原文地址：http://www.cnblogs.com/wqkant/p/5372824.html
