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  • [Leetcode][Python]40: Combination Sum II

    # -*- coding: utf8 -*-
    '''
    __author__ = 'dabay.wang@gmail.com'

    40: Combination Sum II
    https://oj.leetcode.com/problems/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]

    ===Comments by Dabay===
    递归。
    先把candidates排序。
    遍历candidates,用index来记录目前的指针。
    注意去掉结果集中可能的重复。
    '''

    class Solution:
    # @param candidates, a list of integers
    # @param target, integer
    # @return a list of lists of integers
    def combinationSum2(self, candidates, target):
    def combinationSum(candidates, target, index, res, res_list):
    if target == 0:
    if res not in res_list:
    res_list.append(list(res))
    return
    while index < len(candidates) and target >= candidates[index]:
    res.append(candidates[index])
    combinationSum(candidates, target-candidates[index], index+1, res, res_list)
    res.pop()
    index += 1

    res_list = []
    candidates.sort()
    combinationSum(candidates, target, 0, [], res_list)
    return res_list


    def main():
    sol = Solution()
    candidates = [10,1,2,7,6,1,5]
    target = 8
    print sol.combinationSum2(candidates, target)


    if __name__ == "__main__":
    import time
    start = time.clock()
    main()
    print "%s sec" % (time.clock() - start)

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  • 原文地址:https://www.cnblogs.com/Dabay/p/4278232.html
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