[leetcode] 040. Combination Sum II (Medium) (C++)

索引：[LeetCode] Leetcode 題解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode

040. Combination Sum II (Medium) 連結：

題目：https://leetcode.com/problems/combination-sum-ii/
代碼(github)：https://github.com/illuz/leetcode 題意：

跟 039 一樣（給出一些正整數集合，以及一個目標數，從集合中選擇一些數使得它們的和等於目標數），不過不能選重複的數。 分析：

同樣暴力 DFS，不過要考慮重複會複雜點。
還要考慮解集裡不能有相同的解。 代碼：

/**  Author:      illuz <iilluzen[at]gmail.com>*  File:        AC_dfs_n!.cpp*  Create Date: 2015-01-01 11:35:04*  Descripton:  just as the version I*/#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 (rest == 0) {            // to avoid [[1,1,1], 2]            if (!single.empty() && cur < sz && single[single.size() - 1] == candi[cur])                return;            ans.push_back(single);            return;        }        if (sz <= cur || rest < 0)            return;        // choose cur        single.push_back(candi[cur]);        dfs(ans, single, candi, cur + 1, rest - candi[cur]);        single.pop_back();        // don't choose cur        // not contain duplicate combinations        if (!single.empty() && single[single.size() - 1] == candi[cur])            return;        dfs(ans, single, candi, cur + 1, rest);    }public:    vector<vector<int> > combinationSum2(vector<int> &num, int target) {        vector<vector<int> > ans;        vector<int> single;        sort(num.begin(), num.end());        dfs(ans, single, num, 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.combinationSum2(v, tar);    for (auto &i : res) {        for (auto &j : i)            cout << j << ' ';        puts("");    }    return 0;}