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

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

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040. Combination Sum II (Medium)

链接：

题目：https://leetcode.com/problems/combination-sum-ii/
代码(github)：https://github.com/illuz/leetcode

题意：

跟 039 一样（给出一些正整数集合，以及一个目标数，从集合中选择一些数使得它们的和等于目标数），不过不能选重复的数。

分析：

同样暴力 DFS，不过要考虑重复会复杂点。
还要考虑解集里不能有相同的解。

代码：

/*
*  Author:      illuz 
*  File:        AC_dfs_n!.cpp
*  Create Date: 2015-01-01 11:35:04
*  Descripton:  just as the version I
*/

#include 

using namespace std;
const int N = 0;

class Solution {
private:
    void dfs(vector > &ans, vector &single,
            vector &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 > combinationSum2(vector &num, int target) {
        vector > ans;
        vector 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 v(n);
    for (int i = 0; i < n; i++)
        cin >> v[i];
    vector > res = s.combinationSum2(v, tar);
    for (auto &i : res) {
        for (auto &j : i)
            cout << j << ' ';
        puts("");
    }
    return 0;
}