[Leetcode] Combination Sum

描述

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:
All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.
For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[  [7],  [2, 2, 3]]

分析

对于这种求所有组合的题目优先想到用递归的方法解决。方法是：将数组排序，遍历数组，对于数组中每一个元素，考虑“没有被选中”以及“第一个被选中”这两种情况，而每种情况都是原来问题的递归问题。

代码1

class Solution {public:    vector<vector<int>> combinationSum(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) {res.push_back(out); return;}        if (i >= candidates.size() || target < 0) return;        combine(target, candidates, i + 1, out, res);        out.push_back(candidates[i]);        combine(target - candidates[i], candidates, i, out, res);        out.pop_back();    }};

代码2

class Solution {public:    vector<vector<int>> combinationSum(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) {res.push_back(out); return;}        if (i >= candidates.size() || target < 0) return;        for (int j = i; j < candidates.size(); j++) {            out.push_back(candidates[j]);            combine(target - candidates[j], candidates, j, out, res);            out.pop_back();        }    }};

相关题目

Combination Sum II