【leetcode笔记】Python实现 416. Partition Equal Subset Sum

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题目

Given a non-empty array containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal.

Note:

Each of the array element will not exceed 100.
The array size will not exceed 200.

Example 1:

Input: [1, 5, 11, 5]
Output: true

Explanation:

The array can be partitioned as [1, 5, 5] and [11].

Example 2:

Input: [1, 2, 3, 5]
Output: false

Explanation:

The array cannot be partitioned into equal sum subsets.

给定一个数组nums（仅包含正整数），将这个数组切分为两个子数组，使得这两个子数组的和相等。若能完成上述切分，返回True，否则返回False

解析

本题思路，先进行数学分析，设子集为A、B，则有sum(A)+sum(B)=sum(nums),sum(A)=sum(B)，则有sum(A)=sum(nums)/2。故可以将题目转化为从nums中寻找是否有和为sum(nums)/2的子集，有则返回True，否则返回False。

寻找和为sum(nums)/2的子集，可以使用动态规划的思路。用dp[i]来存储和为i的组合个数，对nums里边的数字用n进行遍历，对于所有i>=n的i，有dp[i]=dp[i]+dp[i-n]。

代码

class Solution(object):
    def canPartition(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        # 动态规划
        if nums == []:
            return True
        if sum(nums)%2 == 1:
            return False
        target = sum(nums) // 2
        dp = [0]*(target+1)
        dp[0] = 1
        for n in nums:
            i = target
            while (i>=n):
                dp[i] = dp[i] + dp[i] + dp[i-n]
                i -= 1
        if dp[target] >= 2:
            return True
        else:
            return False
        

一刷：2019年04月05日19:47:59
今天天气真好，清明，周五，下午还游了个泳，耶不耶！！！

参考：https://blog.csdn.net/xiaoxiaoley/article/details/78980823