[LeetCode]385. Mini Parser 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/mini-parser/description/ 题目描述: Given a nested list of integers represented as a string, implement a parser to deserialize it. Each element is either an integer, o…

Question 385. Mini Parser Solution 分析:用NI(count,list)来表示NestedInteger,则解析字符串[123,[456,[789]]]过程如下: # 首先将字符串转化为字符数组,遍历每个字符 [ 压栈操作 NI(0, null) 123 给栈顶元素设置值 NI(0, NI(123)) , 不处理 [ 压栈操作 NI(0, NI(123)) | NI(0, null) 456 给栈顶元素设置值 NI(0, NI(123)) | NI(0, 456…

Given a nested list of integers represented as a string, implement a parser to deserialize it. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Note: You may assume that the string is well-formed: St…

Given a nested list of integers represented as a string, implement a parser to deserialize it. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Note: You may assume that the string is well-formed: St…

Given a nested list of integers represented as a string, implement a parser to deserialize it. Each element is either an integer, or a list -- whose elements may also be integers or other lists. Note: You may assume that the string is well-formed: St…

Given a nested list of integers represented as a string, implement a parser to deserialize it.Each element is either an integer, or a list -- whose elements may also be integers or other lists.Note: You may assume that the string is well-formed:    S…

括号题一般都是stack.. 一开始想的是存入STACK的是SRING,然后POP出括号在构建新的NestedInteger放到另一个里面,但是操作起来费时费力. 后来猛然发现其实可以直接吧NestedInteger作为Object放入Stack里. 这种直接往堆顶元素里放的办法一定要注意. 然后就是edge cases多得一逼,一定要仔细,看了一刷的答案做的,有点后悔.其实有时候觉得麻烦的时候,基本就是思路错了,这个题也是看到一半觉得麻烦,然后发现果然思路错了. public class So…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

Count and Say 思路:递归求出n - 1时的字符串,然后双指针算出每个字符的次数,拼接在结果后面 public String countAndSay(int n) { if(n == 1) return "1"; String front = countAndSay(n - 1); int i = 0; int j = 0; String res = ""; int count = 0; while(j < front.length()){ whi…