Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

violet_program

最新推荐文章于 2025-05-25 18:32:47 发布

阅读量515

点赞数

CC 4.0 BY-SA版权

分类专栏： Recursive

本文链接：https://blog.youkuaiyun.com/violet_program/article/details/8968012

Recursive 专栏收录该内容

30 篇文章

订阅专栏

本文介绍了一种算法，用于从候选数字集合中寻找所有唯一组合，这些组合的数字之和等于目标数字。每个数字只能使用一次，并且解决方案不能包含重复的组合。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations inC where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

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 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]

public class Solution {
    public ArrayList<Integer> cur;
	public ArrayList<ArrayList<Integer>> res;
	public ArrayList<ArrayList<Integer>> combinationSum2(int[] candidates,
			int target) {
		// Start typing your Java solution below
		// DO NOT write main() function
		res = new ArrayList<ArrayList<Integer>>();
		cur = new ArrayList<Integer>();
		Arrays.sort(candidates);
		find(candidates, 0, cur, target);
		return res;
	}
	public void find(int[] candidates, int index, ArrayList<Integer> cur, int target){
		if(target == 0 && !res.contains(cur)){// difference from Combination Sum I
			res.add(new ArrayList<Integer>(cur));
		}
		else{
			for(int i = index; i < candidates.length; i++){
				if(target >= candidates[i]){
					cur.add(candidates[i]);
					find(candidates, i + 1, cur, target - candidates[i]);// and here
					cur.remove(cur.size() - 1);
				}
			}
		}
	}
}

If using DP, I have to count the number of each integers. Too much work.

public class Solution {
	public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
		// IMPORTANT: Please reset any member data you declared, as
		// the same Solution instance will be reused for each test case.
		ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
		ArrayList<Integer> list = new ArrayList<Integer>();
		Arrays.sort(num);
		ArrayList<Integer> value = new ArrayList<Integer>();
		ArrayList<Integer> count = new ArrayList<Integer>();
		int pos = 1;
		int val = num[0];
		for (int i = 1; i < num.length; i++) {
			if (num[i] == val)
				pos++;
			else {
				value.add(val);
				count.add(pos);
				val = num[i];
				pos = 1;
			}
		}
		value.add(val);
		count.add(pos);
		comb(res, list, value, count, 0, num, target);
		return res;
	}

	public void comb(ArrayList<ArrayList<Integer>> res,
			ArrayList<Integer> list, ArrayList<Integer> value,
			ArrayList<Integer> count, int index, int[] num, int target) {
		if (target == 0) {
			res.add(new ArrayList<Integer>(list));
			return;
		}
		int val, quantity;
		for (int i = index; i < value.size(); i++) {
			val = value.get(i).intValue();
			quantity = count.get(i).intValue();
			if (quantity > 0 && target >= val) {
				count.set(i, quantity - 1);
				list.add(val);
				comb(res, list, value, count, i, num, target - val);
				count.set(i, quantity);
				list.remove(list.size() - 1);
			}
		}
	}
}