給定一個無重復元素的正整數(shù)數(shù)組?candidates?和一個正整數(shù)?target?，找出?candidates?中所有可以使數(shù)字和為目標數(shù)?target?的唯一組合。candidates?中的數(shù)字可以無限制重復被選取。如果至少一個所選數(shù)字數(shù)量不同，則兩種組合是唯一的。?對于給定的輸入，保證和為?target 的唯一組合數(shù)少于 150 個。示例?1：輸入: candidates = [2,3,6,7], target = 7 輸出: [[7],[2,2,3]] 示例?2：輸入: candidates = [2,3,5], target = 8 輸出: [[2,2,2,2],[2,3,3],[3,5]] 示例 3：輸入: candidates = [2], target = 1 輸出: [] 示例 4：輸入: candidates = [1], target = 1 輸出: [[1]] 示例 5：輸入: candidates = [1], target = 2 輸出: [[1,1]]鏈接：https://leetcode-cn.com/problems/combination-sum# 方法一 class Solution:def combinateSum(self,condatiate:List,target:int)->List[List[int]]:def dfs(condatitate, begin, size, res, path, target):if target == 0:res.append(path)if target < 0:returnfor index in range(begin, size):dfs(condatitate, index, size, res, path + [condatitate[index]], target - condatitate[index])size = len(condatitate)if size == 0:return []res = []path = []dfs(condatiate, 0, size, res, path, target)return res# 方法二 class Solution:def combinateSum(self,condatiate:List,target:int)->List[List[int]]:def dfs(condatitate, begin, size, res, path, target):if target == 0:res.append(path)for index in range(begin, size):if target - condatitate[index] < 0:breakdfs(condatitate, index, size, res, path + [condatitate[index]], target - condatitate[index])size = len(condatitate)if size == 0:return []res = []path = []condatiate.sort()dfs(condatiate, 0, size, res, path, target)return res

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