Combination Sum

最新推荐文章于 2025-12-18 10:59:28 发布

最新推荐文章于 2025-12-18 10:59:28 发布 · 98 阅读

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本文介绍了一种寻找候选数集合中所有可能组合的算法，这些组合的和等于目标数。文章详细阐述了如何通过递归实现该算法，并确保结果中没有重复的组合，同时保持组合元素的非递减顺序。

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]

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
    	List<List<Integer>> res = new ArrayList<List<Integer>>();  
        List<Integer> item = new ArrayList<Integer>();  
        if (candidates == null || candidates.length==0) {
        	return res;
        }
        Arrays.sort(candidates);
        solve(candidates,target, 0, item ,res);
        return res;
    }

	private void solve(int[] candidates, int target, int start, List<Integer> item,
			List<List<Integer>> res) {
		// TODO Auto-generated method stub
		if (target < 0) {
			return;
		}
		if (target == 0) {
			res.add(new ArrayList<>(item));
			return;
		}
		for (int j = start; j < candidates.length; j++) {
			if (j > 0 && candidates[j] == candidates[j-1]) {
				continue;
			}
			item.add(candidates[j]);
			solve(candidates,target-candidates[j], j, item ,res);
			item.remove(item.size()-1);
		}
	}
}