41. First Missing Positive

/*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.思路：    把每个数字放在正确的位置上，当数字小于数组长度len时。类似桶排序，数字3放在nums[2]的位置。对小于0与大于len的元素不进行处理。当调整完后，a[i]!=i+1时，出现第一个丢失的正整数,返回i+1*/#include <stdio.h>#include <iostream>#include <string>#include <sstream>#include <vector>#include <algorithm>using namespace std;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[nums[i]-1]!=nums[i])swap(nums[i],nums[nums[i]-1]);for(int j=0;j<len;j++)if(nums[j]!=j+1)return j+1;return len+1;    }};int main(){    Solution mys;vector<int> candidates;//={2,3,7};[10, 1, 2, 7, 6, 1, 5]candidates.push_back(1);candidates.push_back(1);candidates.push_back(2);candidates.push_back(7);candidates.push_back(6);candidates.push_back(1);candidates.push_back(5);int res=mys.firstMissingPositive(candidates);cout<<res<<endl;    return 0;}