LeetCode----Combination Sum II

最新推荐文章于 2019-12-20 19:59:10 发布

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本文介绍了一个算法问题“组合总和II”，给定一组候选数字和目标数字，寻找所有不重复的组合，使得候选数字之和等于目标数字。文章提供了Python实现代码及详细解析。

Combination Sum II

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.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
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]

分析：

跟前面的题很相似，给定的candidates中有重复的数，并且每个数只能使用一次。在上一题的代码中稍作修改即可。

代码：

class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        solutions = []
        candidates.sort()
        self.dfs(candidates, target, 0, len(candidates) - 1, list(), solutions)
        return solutions

    def dfs(self, candidates, gap, start, end, knums, solutions):
        if gap == 0:
            if knums not in solutions:
                solutions.append(knums)
            return

        i = start
        while i <= end:
            if gap < candidates[i]:
                return
            self.dfs(candidates, gap - candidates[i], i + 1, end, knums + [candidates[i]], solutions)
            i += 1