【力扣】组合总数（回溯法）

题目描述

#include<iostream>
#include<vector> 
using namespace std;

vector<vector<int> > res;
vector<int> path;

int accumulate(vector<int> path){
	int sum;
	for(int i = 0; i < path.size(); i++){
		sum += path[i];
	}
	return sum;
}
void backtrace(vector<int>& candidates, int target, int startindex){
	//if(accumulate(path.begin(), path.end(),0) == target){
	if(accumulate(path) == target){
		res.push_back(path);
		return ;
	}else if(path.size() == candidates.size()){
		return ;
	}
	
	for(int i = startindex; i < candidates.size(); i++){
		path.push_back(candidates[i]);
		backtrace(candidates, target, i);
		path.pop_back();
	}
}//void 
vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
    res.clear();
	path.clear();
	backtrace(candidates, target, 0);
	return res; 
}
int main(){
	vector<int> candidates;
	int n,m;
	cin>>n;
	while(n--){
		cin>>m;
		candidates.push_back(m);
	}
	int target;
	cin>>target;
	vector<vector<int> > result = combinationSum(candidates, target);
	for(int i = 0; i < result.size(); i++){
		for(int j = 0; j < result[i].size(); j++){
			cout<<result[i][j]<<" ";
		}
		cout<<endl;
	}
	return 0;
}

事实证明纯套模板还是不太可能的，不同的题之间总会有差别，还是要按照三步骤分析，区分同层枚举和向下递归。