标签:
41. First Missing Positive
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Mean:
给你一组整数,找出第一个空缺的正整数.
要求:时间O(n),空间O(n).
analyse:
这题时间O(n)想了半天没想到,用O(n*logn)过的.
然后看了discuss,想法非常巧妙,自愧不如.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-05-18.56
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long(LL);
typedef unsigned long long(ULL);
const double eps(1e-8);
class Solution
{
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target)
{
sort(candidates.begin(),candidates.end());
vector<vector<int>> res;
vector<int> combination;
combinationSum(candidates,res,combination,target,0);
return res;
}
private:
void combinationSum(vector<int>& candidates,vector<vector<int>> &res,vector<int>& combination,int target,int begin)
{
if(!target)
{
res.push_back(combination);
return;
}
for(int i=begin;target>=candidates[i] && i<candidates.size() ;++i)
{
if(i==begin || candidates[i]!=candidates[i-1])
{
combination.push_back(candidates[i]);
combinationSum(candidates,res,combination,target-candidates[i],i+1);
combination.pop_back();
}
}
return;
}
};
int main()
{
freopen("H:\\Code_Fantasy\\in.txt","r",stdin);
int n,target;
while(cin>>n>>target)
{
cout<<n<<" "<<target<<endl;
vector<int> ve;
for(int i=0; i<n; ++i)
{
int tmp;
cin>>tmp;
ve.push_back(tmp);
}
Solution solution;
vector<vector<int>> ans=solution.combinationSum2(ve,target);
for(auto p1:ans)
{
for(auto p2:p1)
{
cout<<p2<<" ";
}
cout<<endl;
}
}
return 0;
}
/*
*/
O(n)时间,O(1)空间的代码:
class Solution {
public:
int firstMissingPositive(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n; i++)
while (nums[i] > 0 && nums[i] <= n && nums[nums[i] - 1] != nums[i])
swap(nums[i], nums[nums[i] - 1]);
for (int i = 0; i < n; i++)
if (nums[i] != i + 1)
return i + 1;
return n + 1;
}
};
LeetCode - 41. First Missing Positive
标签:
原文地址:http://www.cnblogs.com/crazyacking/p/5246043.html