[Leetcode][Python]39: Combination Sum

本文介绍了解决LeetCode第39题“组合总和”的递归算法。该算法首先对候选数进行排序,然后通过递归调用寻找所有可能的组合,这些组合的元素之和等于目标值。文章详细解释了递归函数的实现过程,并提供了完整的Python代码示例。
# -*- coding: utf8 -*-
'''
__author__ = 'dabay.wang@gmail.com'

39: Combination Sum
https://oj.leetcode.com/problems/combination-sum/

Given a set of candidate numbers (C) and a target number (T),
find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.

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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

===Comments by Dabay===
递归。
先把candidates排序。
遍历candidates,用index来记录目前的指针。
'''

class Solution:
# @param candidates, a list of integers
# @param target, integer
# @return a list of lists of integers
def combinationSum(self, candidates, target):
def combinationSum2(candidates, target, index, res, res_list):
if target == 0:
res_list.append(list(res))
return
while index < len(candidates) and target >= candidates[index]:
res.append(candidates[index])
combinationSum2(candidates, target-candidates[index], index, res, res_list)
res.pop()
index += 1

res_list = []
candidates.sort()
combinationSum2(candidates, target, 0, [], res_list)
return res_list


def main():
sol = Solution()
candidates = [1,2]
target = 4
print sol.combinationSum(candidates, target)


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

转载于:https://www.cnblogs.com/Dabay/p/4278231.html

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