Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…
Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…
[抄题]: Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may ass…
Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…
Leetcode 91. Decode Ways 解码方法(动态规划,字符串处理) 题目描述 一条报文包含字母A-Z,使用下面的字母-数字映射进行解码 'A' -> 1 'B' -> 2 ... 'Z' -> 26 给一串包含数字的加密报文,求有多少种解码方式 举个例子,已知报文"12",它可以解码为AB(1 2),也可以是L (12) 所以解码方式有2种. 测试样例 Input: "0" "121212" "1010…
给定一个经过编码的字符串,返回它解码后的字符串.编码规则为: k[encoded_string],表示其中方括号内部的 encoded_string 正好重复 k 次.注意 k 保证为正整数.你可以认为输入字符串总是有效的:输入字符串中没有额外的空格,且输入的方括号总是符合格式要求的.此外,你可以认为原始数据不包含数字,所有的数字只表示重复的次数 k ,例如不会出现像 3a 或 2[4] 的输入.示例:s = "3[a]2[bc]", 返回 "aaabcbc".s…
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 栈 日期 题目地址:https://leetcode.com/problems/decode-string/description/ 题目描述 Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the…
题目描述:按照规定,把字符串解码,具体示例见题目链接 思路:使用两个栈分别存储数字和字母 注意1: 数字是多位的话,要处理后入数字栈 注意2: 出栈时过程中产生的组合后的字符串要继续入字母栈 注意3: 记得字母出栈的时候字符要逆序组合成字符串 注意4: 不用字符串而用字母栈的原因是字符串的 join 效率会比字符串加法高一些 结果: 30 ms, beat 98.02% 缺点:判断是数字那里有点代码不简洁,可以把 j 挪到循环外面的 class Solution(object): def dec…
题目如下: 解题思路:这种题目和四则运算,去括号的题目很类似.解法也差不多. 代码如下: class Solution(object): def decodeString(self, s): """ :type s: str :rtype: str """ stack = [] for i in s: if i != ']': stack.append(i) continue repeatStr = '' while len(stack) >…
A message containing letters from A-Z is being encoded to numbers using the following mapping: 'A' -> 1 'B' -> 2 ... 'Z' -> 26 Given a non-empty string containing only digits, determine the total number of ways to decode it. Example 1: Input: &qu…