【leetcode】【medium】39. Combination Sum

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本文深入探讨了LeetCode上的组合总和问题，采用回溯法高效求解，通过剪枝和去重策略优化算法，避免重复组合，适用于寻找候选数集合中所有可能的组合，使这些数的和等于目标值。

39. Combination Sum

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 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]
]

题目链接：https://leetcode-cn.com/problems/combination-sum/

思路

回溯法，向下传递结果运行效率更高。

剪枝：对候选人做排序，这样每一次选的都是数组里最小的数值，若比目标还大，则需要剪枝。

去重：对于组合的题目，需要对同一组合不同排序的结果去重，方法就是在组合答案时只看>=自身的数字。

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