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。这里背包容量为total_sum/2 ...
.Last Stone Weight II https: leetcode.com problems last stone weight ii 题意:从一堆石头里任选两个石头s ,s ,若s s ,则两个石头都被销毁,否则加入s lt s ,剩下一块重量为s s 的石头。重复上面的操作,直至只剩一块石头,或没有石头。问最后剩下的石头的重量最小为多少 表示没有剩余石头 。 解法: 每次选两块石头进 ...
2019-05-19 18:08 0 507 推荐指数:
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。这里背包容量为total_sum/2 ...
You are given an array of integers stones where stones[i] is the weight of the ith stone. We are playing a game with the stones. On each turn, we ...
原题链接在这里:https://leetcode.com/problems/last-stone-weight/ 题目: We have a collection of rocks, each rock has a positive integer weight. Each turn, we ...
We have a collection of stones, each stone has a positive integer weight. Each turn, we choose the two heaviest stones and smash them together. ...
/strengthen/LeetCode➤原文地址:https://www.cnblogs.com/streng ...
Alice and Bob continue their games with piles of stones. There are a number of piles arranged in ...
There is a stone game.At the beginning of the game the player picks n piles of stones in a circle. The goal is to merge the stones in one ...
Given a nested list of integers, return the sum of all integers in the list weighted by their dep ...