#yyds干货盘点# leetcode算法题：组合总和 II

Posted 2022-06-30 灰太狼_cxh

题目：

给定一个候选人编号的集合 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。

candidates 中的每个数字在每个组合中只能使用 一次 。

注意：解集不能包含重复的组合。 

 

示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,

输出:

[

[1,1,6],

[1,2,5],

[1,7],

[2,6]

]

示例 2:

输入: candidates = [2,5,2,1,2], target = 5,

输出:

[

[1,2,2],

[5]

]

代码实现：

class Solution 
    List<int[]> freq = new ArrayList<int[]>();
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    List<Integer> sequence = new ArrayList<Integer>();

    public List<List<Integer>> combinationSum2(int[] candidates, int target) 
        Arrays.sort(candidates);
        for (int num : candidates) 
            int size = freq.size();
            if (freq.isEmpty() || num != freq.get(size - 1)[0]) 
                freq.add(new int[]num, 1);
             else 
                ++freq.get(size - 1)[1];
            
        
        dfs(0, target);
        return ans;
    

    public void dfs(int pos, int rest) 
        if (rest == 0) 
            ans.add(new ArrayList<Integer>(sequence));
            return;
        
        if (pos == freq.size() || rest < freq.get(pos)[0]) 
            return;
        

        dfs(pos + 1, rest);

        int most = Math.min(rest / freq.get(pos)[0], freq.get(pos)[1]);
        for (int i = 1; i <= most; ++i) 
            sequence.add(freq.get(pos)[0]);
            dfs(pos + 1, rest - i * freq.get(pos)[0]);
        
        for (int i = 1; i <= most; ++i) 
            sequence.remove(sequence.size() - 1);