LeetCode-039 Combination Sum

最新推荐文章于 2021-03-18 11:37:34 发布

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本文介绍了一种使用深度优先搜索(DFS)解决从给定数组中寻找所有可能的组合使其和等于特定目标值的问题的方法。通过排序输入数组进行剪枝优化，有效避免了重复组合。

Description

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

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

Note

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

Example

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]

Analyse

题意为让你在给定的正整数数组里面找到和为目标值的组合并输出，其中数字可以重复使用，看起来就是一道很DFS的题，所以就用DFS来做了。在做DFS的时候，可以先排序来进行一些剪枝，会更快一点。

Code

class Solution {
public:
    void calcombination(vector<vector<int>> &ans, vector<int> &temp, int target, vector<int>& candidates, int begin) {
        if (target == 0) {
            ans.push_back(temp);
        } else {
            for (int i = begin; i < candidates.size(); i++) {
                if (candidates[i] > target) break;
                temp.push_back(candidates[i]);
                calcombination(ans, temp, target - candidates[i], candidates, i);
                temp.pop_back();
            }
        }
    }
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        vector<vector<int>> ans;
        vector<int> temp;
        calcombination(ans, temp, target, candidates, 0);
        return ans;
    }
};