题目描述: You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope. Wh…

https://leetcode.com/problems/russian-doll-envelopes/ You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater tha…

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope. What is…

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope. What is…

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.What is t…

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope. What is…

Leetcode354 暴力的方法是显而易见的 O(n^2)构造一个DAG找最长链即可. 也有办法优化到O(nlogn) 注意 信封的方向是不能转换的. 对第一维从小到大排序,第一维相同第二维从大到小排序. 维护一个符合题意的队列,当队列中的第二维均比当前信封的第二维小时,必然可以增加到队尾. 如果不然,可以让当前信封作为“替补”,它可以在恰当的时候代替恰好比它大的信封. 当替补们足够替换所有已有信封时,就可以增加新的信封了. 比较抽象,不过这个技巧很有趣. 看代码吧,很清晰. class So…

[4]Median of Two Sorted Arrays [29]Divide Two Integers [33]Search in Rotated Sorted Array [34]Find First and Last Position of Element in Sorted Array [35]Search Insert Position [50]Pow(x, n) [69]Sqrt(x) [74]Search a 2D Matrix (2019年1月25日,谷歌tag复习)剑指of…

[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 递归和非递归,此提比较简单.广度优先遍历即可.关键之处就在于如何保持访问深度. 下面是4种代码: im…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. Note: Could you optimize your algorithm to use only O(k) extra space? 思路:最简单的方法就是依照[Leetcode]Pascal's Triangle 的方式自顶向下依次求解,但会造成空间的浪费.若仅仅用一个vect…