leetcode 1049 Last Stone Weight II(最后一塊石頭的重量 II)
思路：原問題可以轉換為將數組分割為兩個集合(根據符號為正和符號為負划分)，使得這兩個集合和的差最小。 可以等價為01背包問題。那么dp[i][j]就是將前i個物品放到容量為j的背包能得到的最大值。 ...
原文：[LeetCode] 1049. Last Stone Weight II 最后的石頭重量之二
You are given an array of integersstoneswherestones i is the weight of theithstone. We are playing a game with the stones. On each turn, we choose any two stones and smash them together. Suppose the ...
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