leetcode_1049. Last Stone Weight II_[DP]】的更多相关文章

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…
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. …
Last Stone Weight II 欢迎关注H寻梦人公众号 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 choose any two stones and smash them together. Suppose the stones have…
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/…
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…
problem 1046. Last Stone Weight 参考 1. Leetcode_easy_1046. Last Stone Weight; 完…
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…
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A Famous Stone Collector Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 793    Accepted Submission(s): 292 Problem Description Mr. B loves to play with colorful stones. There are n colors of…
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…