LeetCode 39. Combination Sum

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

复制Input: candidates = [2,3,6,7], target = 7
Output: [[2,2,3],[7]]
Explanation:
2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
7 is a candidate, and 7 = 7.
These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8
Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1
Output: []

Example 4:

Input: candidates = [1], target = 1
Output: [[1]]

Example 5:

Input: candidates = [1], target = 2
Output: [[1,1]]

Constraints:

1 <= candidates.length <= 30
1 <= candidates[i] <= 200
All elements of candidates are distinct.
1 <= target <= 500

实现思路：

题意就是组合数之和等于目标数，要求输出所有组合可能，一提到“组合”,应该物理反应就是回溯算法，本题不过多赘述在回溯标签里总结了大量回溯算法题。

AC代码代码

class Solution {
		vector<int> temp;
		vector<vector<int>> ans;
		void dfs(int idx,int sum,int target,int k,vector<int> &sq) {
			if(sum==target) {
				ans.push_back(temp);
				return;
			}
			if(sum>target||idx>=k) return;//剪枝 当数的总和已经大于目标数或者当数组索引下标超出范围的时候就退出
			if(sq[idx]+sum<=target) {
				temp.push_back(sq[idx]);
				dfs(idx,sum+sq[idx],target,k,sq);
				temp.pop_back();
				dfs(idx+1,sum,target,k,sq);
			}
		}

	public:
		vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
			sort(candidates.begin(),candidates.end());
			dfs(0,0,target,candidates.size(),candidates);
			return ans;
		}
};