leetcode 1049 Last Stone Weight II(最后一块石头的重量 II)
思路：原问题可以转换为将数组分割为两个集合(根据符号为正和符号为负划分)，使得这两个集合和的差最小。 可以等价为01背包问题。那么dp[i][j]就是将前i个物品放到容量为j的背包能得到的最大值。 ...
原文：[LeetCode] 1049. Last Stone Weight II 最后的石头重量之二
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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