leetcode--Combination Sum

最新推荐文章于 2020-05-26 21:39:27 发布

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本文介绍了一种算法，用于在给定的候选数集中找到所有可能的组合，使得这些组合的元素之和等于指定的目标数。算法允许重复选择候选数，并确保组合中的数按非降序排列，避免重复结果。

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, … , a_k) must be in non-descending order. (ie, a₁ ≤ a₂ ≤ … ≤ a_k).
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]

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
    	List<List<Integer>> res = new ArrayList<List<Integer>>();
    	Arrays.sort(candidates);
    	Stack<Integer> stack = new Stack<Integer>();
    	dfs(res, stack, target, 0, candidates,0);
    	return res;
    }
    
    void dfs(List<List<Integer>> res,Stack<Integer> stack,int target,int cur,int[] arr,int sum){    	
    	if(sum==target){
    		ArrayList<Integer> data = new ArrayList<Integer>(stack);
    		res.add(data);
    		return;
    	}else if(sum>target) return;
    	else{
    		for(int i=cur;i<arr.length;i++){
    			stack.add(arr[i]);
    			sum += arr[i];
    			dfs(res, stack, target, i, arr,sum);
    			sum -= arr[i];
    			stack.pop();
    		}
    	}
    }
}