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33 lines
1.2 KiB
Python
33 lines
1.2 KiB
Python
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#coding: utf-8
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''' mbinary
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#######################################################################
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# File : last-stone-weight.py
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# Author: mbinary
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# Mail: zhuheqin1@gmail.com
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# Blog: https://mbinary.xyz
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# Github: https://github.com/mbinary
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# Created Time: 2019-05-28 23:30
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# Description:
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leetcode 1049: https://leetcode-cn.com/problems/last-stone-weight-ii/
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有一堆石头，每块石头的重量都是正整数。
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每一回合，从中选出任意两块石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
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如果 x == y，那么两块石头都会被完全粉碎；
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如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
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最后，最多只会剩下一块石头。返回此石头最小的可能重量。如果没有石头剩下，就返回 0。
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#######################################################################
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'''
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class Solution:
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def lastStoneWeightII(self, stones: List[int]) -> int:
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sm = sum(stones)
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ans = sm//2
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dp = [0]*(ans+1)
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for x in stones:
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for j in range(ans,x-1,-1):
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dp[j] = max(dp[j],dp[j-x]+x)
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return sm-2*dp[ans]
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