[leetcode] Combination Sum-优快云博客

本文探讨了LeetCode上的经典问题“组合总和”。通过递归算法寻找候选数集中所有可能的组合，使得这些数的和等于目标数。文章提供了一个C++实现方案，并详细解释了如何避免重复组合。

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From : https://leetcode.com/problems/combination-sum/

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:
[7]
[2, 2, 3]

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<vector<int>> ans;
        if(candidates.size()==0 || target <=0 ) return ans;
        sort(candidates.begin(), candidates.end());
        vector<int> cur;
        find(candidates, cur, ans, candidates.size()-1, target);
        return ans;
    }
    
        void find(vector<int>& cands, vector<int>& cur, vector<vector<int>>& ans, int index, int need) {
        if(need == 0) ans.push_back(cur);
        if(index < 0) return;
        for(int i=index; i>=0; i--) {
            if(need >= cands[i]) {
				cur.insert(cur.begin(), cands[i]);
                find(cands, cur, ans, i, need-cands[i]);
				cur.erase(cur.begin());
            }
        }
    }
};