[LeetCode]Combination Sum

题目

Number: 35

Difficulty: Medium

Tags: Array, Backtracking

Given a set of candidate numbers (C) 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.

Elements in a combination

(a1, a2, … , ak)

must be in non-descending order.

(ie, a1 ≤ a2 ≤ … ≤ ak)

.

The solution set must not contain duplicate combinations.

For example, given candidate set

2,3,6,7

and target

7

,

A solution set is:

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

题解

在数组中寻找和为目标值的所有组合。

典型的回溯问题。

代码

[code]class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int> > res;
        vector<int> combination;
        combinationSum(candidates, target, res, combination, 0);
        return res;
    }
private:
    void combinationSum(vector<int> &candidates, int target, vector<vector<int> > &res, vector<int> &combination, int begin) {
        if  (!target) {
            res.push_back(combination);
            return;
        }
        for (int i = begin; i != candidates.size() && target >= candidates[i]; ++i) {
            combination.push_back(candidates[i]);
            combinationSum(candidates, target - candidates[i], res, combination, i);
            combination.pop_back();
        }
    }
};