LeetCode 40. 组合总和 II

Posted 2023-06-08 穿过雾的阴霾

class Solution 
public:
    vector<vector<int>> res;
    vector<vector<int>> combinationSum2(vector<int>& candidates, int target) 
        sort(candidates.begin(),candidates.end());
        dfs(candidates,0,target);
        return res;
    
    vector<int> path;
    void dfs(vector<int>& candidates, int u,int target)//枚举每个数选或不选
    
        if(u>=candidates.size()||target<=0)
        
            if(!target) res.push_back(path);
            return;
        
        path.push_back(candidates[u]);
        dfs(candidates,u+1,target-candidates[u]);
        path.pop_back();
        //不选
        int idx=u;
        while(idx<candidates.size()&&candidates[idx]==candidates[u])   idx++;
        dfs(candidates,idx,target);
    
;

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LeetCode 40.组合总和II

给定一个数组?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。

说明：

所有数字（包括目标数）都是正整数。
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,
所求解集为:
[
[1, 7],
[1, 2, 5],
[2, 6],
[1, 1, 6]
]
示例?2:

输入: candidates =?[2,5,2,1,2], target =?5,
所求解集为:
[
? [1,2,2],
? [5]
]

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        ans = []
        cur = []
        candidates = sorted(candidates)
        self.dfs(candidates,target,ans,cur,0)
        return ans
    def dfs(self,candidates,target,ans,cur,s):
        if 0 == target:
            ans.append(cur[::])
            return
        for i in range(s,len(candidates)):
            #print(‘ans:‘,ans,‘cur:‘,cur,‘s:‘,s)
            num = candidates[i]
            if num > target:
                break
            else:
                if i>s and candidates[i] == candidates[i-1]:
                    continue
                cur.append(num)
                self.dfs(candidates,target-num,ans,cur,i+1)
                cur.pop()