40. Combination Sum II

最新推荐文章于 2025-05-25 18:32:47 发布

原创最新推荐文章于 2025-05-25 18:32:47 发布 · 181 阅读

0 ·

CC 4.0 BY-SA版权

刷题日记专栏收录该内容

51 篇文章

订阅专栏

本文介绍了一个算法问题：给定一组候选数字和一个目标值，在候选数字中找出所有可能的组合，使得这些数字之和等于目标值。每个数字在每种组合中只能使用一次，并且解决方案不能包含重复的组合。

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C 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.
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]
]

这一题和上一题不一样的是，已经遍历过的元素只能出现一次

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        ArrayList<List<Integer>> finans = new ArrayList<List<Integer>>();
        ArrayList<Integer> curans = new ArrayList<Integer>();
        if(candidates.length==0||candidates==null) return finans;
        Arrays.sort(candidates);
        boolean[] visited = new boolean[candidates.length];
        buildAns(candidates,target,0,curans,finans,visited);
        return finans;
    }
    
    private void buildAns(int[] candidates, int target, int start, ArrayList<Integer> curans, ArrayList<List<Integer>> finans,boolean[] visited){
        if(target==0){
            finans.add(new ArrayList<Integer> (curans));
            return;
        }
        if(target<0)
            return;
        for(int i=start;i<candidates.length;i++){
            if(!visited[i]){
                if(i>0 && candidates[i]==candidates[i-1] && visited[i-1]==false) continue;
                curans.add(candidates[i]);
                visited[i]=true;
                int newtarget=target-candidates[i];
                buildAns(candidates,newtarget,i+1,curans,finans,visited);
                visited[i]=false;
                curans.remove(curans.size()-1);
            }
        }
    }
}