• Leetcode** 39. Combination Sum


    Description: Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order.

    The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different.

    It is guaranteed that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

    Link: 39. Combination Sum

    Examples:

    Example 1:
    Input: candidates = [2,3,6,7], target = 7
    Output: [[2,2,3],[7]]
    Explanation:
    2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times.
    7 is a candidate, and 7 = 7.
    These are the only two combinations.
    
    Example 2:
    Input: candidates = [2,3,5], target = 8
    Output: [[2,2,2,2],[2,3,3],[3,5]]
    
    Example 3:
    Input: candidates = [2], target = 1
    Output: []
    
    Example 4:
    Input: candidates = [1], target = 1
    Output: [[1]]
    
    Example 5:
    Input: candidates = [1], target = 2
    Output: [[1,1]]

    思路: 返回所有可能的组合,其和为target,所有大可能要用回溯法。dfs遍历的过程中返回所有路径,每条路径的终止条件是target == 0. append(path),target < 0,说明这条搜索的路径不可行,return但不添加到新路径. 值得注意的是,因为每个数字是可以无限次被挑选的,所以再次进入dfs(),start_index依旧是i,不是i+1.

    class Solution(object):
        def combinationSum(self, candidates, target):
            """
            :type candidates: List[int]
            :type target: int
            :rtype: List[List[int]]
            """
            self.res = []
            self.dfs(candidates, 0, target, [])
            return self.res
        
        def dfs(self, candidates, start, target, path):
            if target < 0:
                return
            if target == 0:
                self.res.append(path)
                return
            for i in range(start, len(candidates)):
                self.dfs(candidates, i, target-candidates[i], path+[candidates[i]])

    当然也可以先对candidates排序,这样就可以提早break,终止搜索,会节省时间。

    class Solution(object):
        def combinationSum(self, candidates, target):
            """
            :type candidates: List[int]
            :type target: int
            :rtype: List[List[int]]
            """
            self.res = []
            candidates.sort()
            self.dfs(candidates, 0, target, [])
            return self.res
        
        def dfs(self, candidates, start, target, path):
            if target < 0:
                return
            if target == 0:
                self.res.append(path)
                return
            for i in range(start, len(candidates)):
                if target < candidates[i]:
                    break
                self.dfs(candidates, i, target-candidates[i], path+[candidates[i]])

    日期: 2021-04-21  谷雨便已是过了。

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  • 原文地址:https://www.cnblogs.com/wangyuxia/p/14684013.html
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