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 choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and y with x…

lc57 Insert Interval 仔细分析题目,发现我们只需要处理那些与插入interval重叠的interval即可,换句话说,那些end早于插入start以及start晚于插入end的interval都可以保留.我们只需要两个指针,i&j分别保存重叠interval中最早start和最晚end即可,然后将interval [i, j]插入即可 class Solution { public int[][] insert(int[][] intervals, int[] newInte…

1046. 最后一块石头的重量 1046. Last Stone Weight 题目描述 每日一算法2019/6/22Day 50LeetCode1046. Last Stone Weight Java 实现 and so on 参考资料 https://leetcode.com/problems/last-stone-weight/ https://leetcode-cn.com/problems/last-stone-weight/…

problem 1046. Last Stone Weight 参考 1. Leetcode_easy_1046. Last Stone Weight; 完…

原题链接在这里:https://leetcode.com/problems/last-stone-weight-ii/ 题目: We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose any two rocks and smash them together.  Suppose the stones have weights x and y with x <= y. …

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 大根堆 日期 题目地址:https://leetcode.com/problems/last-stone-weight/ 题目描述 We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose the two heaviest…

题目如下: We have a collection of rocks, each rock has a positive integer weight. Each turn, we choose the two heaviest rocks and smash them together.  Suppose the stones have weights x and y with x <= y.  The result of this smash is: If x == y, both sto…

有一堆石头,每块石头的重量都是正整数. 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: 如果 x == y,那么两块石头都会被完全粉碎: 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. 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. 示例: 输入:[2,7,4,1,8,1] 输出:1 解释: 组合 2 和 4,得到 2,所以数组转…

1049. 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s1,s2,若s1==s2,则两个石头都被销毁,否则加入s1<s2,剩下一块重量为s2-s1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(0表示没有剩余石头). 解法: 每次选两块石头进行相减问最后一块石头重量最小为多少,可以看作是将所有石头分为两波,使两波石头的差值的…

Given a nested list of integers, return the sum of all integers in the list weighted by their depth. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Different from the [leetcode]339. Nested List Wei…