[Leetcode] Combination Sum II

描述

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:

[  [1, 7],  [1, 2, 5],  [2, 6],  [1, 1, 6]]

分析

这道题是 Combination Sum 的改进版，与原题不同的地方在于这道题要求数组中的每一个元素只能使用一次。

因此还是使用原来题目的思路，还是使用递归的方法来解决题目，原题的求解看 这里 。

代码1

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int>> res; vector<int> out;        combine(target, candidates, 0, out, res);        return res;    }    void combine(int target, vector<int>& candidates, int i, vector<int>& out, vector<vector<int>>& res) {       if (target < 0) return;       if (target == 0) {res.push_back(out); return;}       for (int j = i; j < candidates.size(); j++) {           if (j > i && candidates[j] == candidates[j - 1]) continue;           out.push_back(candidates[j]);           combine(target - candidates[j], candidates, j + 1, out, res);           out.pop_back();       }    }};

代码2

class Solution {public:    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {        sort(candidates.begin(), candidates.end());        vector<vector<int>> res; vector<int> out;        combine(target, candidates, 0, out, res);        return res;    }    void combine(int target, vector<int>& candidates, int i, vector<int>& out, vector<vector<int>>& res) {       if (target < 0) return;       if (target == 0) {res.push_back(out); return;}       for (int j = i; j < candidates.size(); j++) {           if (j > i && candidates[j] == candidates[j - 1]) continue;           out.push_back(candidates[j]);           combine(target - candidates[j], candidates, j + 1, out, res);           out.pop_back();       }    }};

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Combination Sum