[LeetCode] 039. Combination Sum (Medium) (C ++), combinationleetcode

[LeetCode] 039. Combination Sum (Medium) (C ++), combinationleetcode

Index: [LeetCode] Leetcode index (C ++/Java/Python/SQL)
Github: https://github.com/illuz/leetcode

039. Combination Sum (Medium) Link:

Title: https://leetcode.com/problems/combination-sum/
Code (github): https://github.com/illuz/leetcode

Question:

A positive integer set and a target number are given. You can select a number from the set so that their sum is equal to the target number.
The resulting solution set cannot be repeated.

Analysis:

Brute-force search is acceptable in the past. First, sort the order and then use DFS. There are two options each time. One is to select the current number and then recursion the current number, second, if the current number is not selected, the current number is recursive.
It is good to understand DFS.

Code(C ++)

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        AC_dfs_n!.cpp*  Create Date: 2015-01-01 10:45:58*  Descripton:  dfs, choose or not choose*/#include <bits/stdc++.h>using namespace std;const int N = 0;class Solution {private:    void dfs(vector<vector<int> > &ans, vector<int> &single,            vector<int> &candi, int cur, int rest) {        int sz = candi.size();        if (sz <= cur || rest < 0)            return;        if (rest == 0) {            ans.push_back(single);            return;        }        // choose cur        single.push_back(candi[cur]);        dfs(ans, single, candi, cur, rest - candi[cur]);        single.pop_back();        // don't choose cur        dfs(ans, single, candi, cur + 1, rest);    }public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        vector<vector<int> > ans;        vector<int> single;        sort(candidates.begin(), candidates.end());        dfs(ans, single, candidates, 0, target);        return ans;    }};int main() {    int tar;    int n;    Solution s;    cin >> n >> tar;    vector<int> v(n);    for (int i = 0; i < n; i++)        cin >> v[i];    vector<vector<int> > res = s.combinationSum(v, tar);    for (auto &i : res) {        for (auto &j : i)            cout << j << ' ';        puts("");    }    return 0;}

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