Leetcode 40.组合总和II

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40.组合总和II

题目

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。

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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]
]

方法

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class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res = []
        vis = set()
        def dfs(path, start, target):
            if target == 0:
                res.append(path[:])
                return

            for i in range(start, len(candidates)):
                if i >start and candidates[i] == candidates[i-1]:
                    continue
                if not i in vis and target - candidates[i] >= 0:
                    vis.add(i)
                    path.append(candidates[i])
                    dfs(path, i+1, target - candidates[i])
                    path.pop()
                    vis.discard(i)

        candidates.sort()
        dfs([], 0, target)
        return res
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class Solution(object):
    def combinationSum2(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """

        def combine(candidates, target, p, ans):
            if target == 0:
                ans.append(p)
                return

            for i, c in enumerate(candidates):
                if i > 0 and c == candidates[i-1]:
                    continue
                if target - c < 0:
                    break
                combine(candidates[i+1:], target - c, p + [c], ans)
        ans = []
        candidates.sort()
        combine(candidates, target, [], ans)
        return ans