Combination Sum (Java)

最新推荐文章于 2021-03-04 10:34:23 发布

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最新推荐文章于 2021-03-04 10:34:23 发布

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本文链接：https://blog.youkuaiyun.com/tofu_jelly/article/details/42871191

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本文详细介绍了如何使用递归方法解决在一组候选数中找到所有可能的组合，使得这些组合的元素之和等于特定目标数的问题。通过排序、回溯和剪枝等策略，有效地避免了重复解的产生，确保了解答的唯一性和效率。

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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]

注意，但凡不出现重复[2,2,3] [3,2,2]这种情况，都要设一个i的start传入递归。这道题的解决方法和题目subset以及combinations类似。

Source

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> st = new ArrayList<List<Integer>>();
        List<Integer> a = new ArrayList<Integer>();
    	if(candidates.length == 0) return st;
    	int sum = 0;
    	int start = 0;
    	
    	Arrays.sort(candidates);  //这道题的测试数据有乱序的
    	dfs(start, sum, candidates, target, st, a);
    	return st;
    }
    public void dfs(int start, int sum, int[] candidates, int target, List<List<Integer>> st, List<Integer> a){
    	if(sum > target) return;
    	if(sum == target){
    		st.add(new ArrayList<Integer>(a));
    		return;
    	}
    	
    	for(int i = start; i < candidates.length; i++){
    		a.add(candidates[i]);
    		dfs(i, sum + candidates[i], candidates, target, st, a); //注意sum是要随着dfs变化的
    		a.remove(a.size() - 1);
    	}
    }
}

Test

    public static void main(String[] args){
    	int[] candidates = {2,3,6,7};
    	int target = 7;
    	System.out.println(new Solution().combinationSum(candidates, target));
    
    }