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。這里背包容量為total_sum/2 ...
原文：leetcode_1049. Last Stone Weight II_[DP]
.Last Stone Weight II https: leetcode.com problems last stone weight ii 題意：從一堆石頭里任選兩個石頭s ,s ，若s s ,則兩個石頭都被銷毀，否則加入s lt s ，剩下一塊重量為s s 的石頭。重復上面的操作，直至只剩一塊石頭，或沒有石頭。問最后剩下的石頭的重量最小為多少 表示沒有剩余石頭 。 解法： 每次選兩塊石頭進 ...
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