[LeetCode] 416. Partition Equal Subset Sum_Medium tag: backpack

标签：sub   size   integer   xpl   example   pos   self   empty   set   

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:

1. Each of the array element will not exceed 100.
2. 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.

这个实际上就是0，1背包的一个马甲题，可以看作背包的size是sum(nums) //2 如果能够被平分的话。

用dp[n + 1][背包size + 1] 去初始化dp

dp[i][j] = dp[i - 1][j] or dp[i - 1][j - nums[i - 1]] if j >= nums[i - 1] else dp[i - 1][j]     ( i = [1, n], j = [1, 背包size])

initial:

dp[i][0] = 0    (i = [0, n])

T: O(n * sum(nums))    S: O(n * sum(nums))

Code

class Solution:
    def canPartition(self, nums):
        n, s = len(nums), sum(nums)
        if s % 2: return False
        target = s // 2
        dp = [[False] * (target + 1) for _ in range(n + 1)]
        for i in range(n + 1):
            dp[i][0] = True
        for i in range(1, n + 1):
            for j in range(1, target + 1):
                dp[i][j] = dp[i - 1][j] or dp[i - 1][j - nums[i - 1]] if j >= nums[i - 1] else dp[i - 1][j]
            if j == target and dp[i][j]:
                return True
        return False

可用滚动数组将Space 优化为O(sum(nums))

[LeetCode] 416. Partition Equal Subset Sum_Medium tag: backpack

标签：sub   size   integer   xpl   example   pos   self   empty   set   

原文地址：https://www.cnblogs.com/Johnsonxiong/p/11136141.html