字符串压缩。利用字符重复出现的次数,编写一种方法,实现基本的字符串压缩功能。比如,字符串aabcccccaaa会变为a2b1c5a3。若“压缩”后的字符串没有变短,则返回原先的字符串。你可以假设字符串中只包含大小写英文字母(a至z)。 代码语言: 代码 输入:"abbccd"输出:"abbccd"解释:"abbccd"压缩后为"a1b2c2d1",...
力扣leetcode-cn.com/problems/basic-calculator/ 题目描述 给你一个字符串表达式 s ,请你实现一个基本计算器来计算并返回它的值。 示例1: 输入:s = "1 + 1" 输出:2 示例2: 输入:s = " 2-1 + 2 " 输出:3 示例3: 输入:s = "(1+(4+5+2)-3)+(6+8)" 输出:23提示...
LeetCode224 Hard 基本计算器https://leetcode-cn.com/problems/basic-calculator/ LeetCode1006 Medium 笨阶乘https://leetcode-cn.com/problems/clumsy-factorial/ (3)字符串类 LeetCode394 Medium 字符串解码https://leetcode-cn.com/problems/decode-string/ 2.单调栈 LeetCode316 Medium 去除重复字母https:/...
classSolution {public:intcalculate(strings) {intres =0, num =0, sign =1, n =s.size();for(inti =0; i < n; ++i) {charc =s[i];if(c >='0'&& c <='9') { num=10* num + (c -'0'); }elseif(c =='(') {intj = i, cnt =0;for(; i < n; ++i) {if(s[i] ==...
1//The main idea of this is the left bracket might change the sign of a number, however, this does not seem to be very generalized2//https://leetcode.com/problems/basic-calculator/discuss/62362/JAVA-Easy-Version-To-Understand!!!3publicclassSolution {4//1-2-(4+5+2)-35publicintcalcu...
227 Basic Calculator II 22.2% Medium Minimum Height Trees 【题目】For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs)....
772.Basic-Calculator-III (H) 385.Mini-Parser (H) 439.Ternary-Expression-Parser (H-) 591.Tag-Validator (H) 726.Number-of-Atoms (M+) 1087.Brace-Expansion (H) 1096.Brace-Expansion-II (H) 1106.Parsing-A-Boolean-Expression (H-) 1896.Minimum-Cost-to-Change-the-Final-Value-of-Expression...
id: fuxuemingzhu 个人博客:http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/basic-calculator-ii/description/ 题目描述: Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators and empty spa...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算法解决。大家加油!:) 0 stars 720 forks ...
把string转换成数字。atoi(s.substr(i, j - i).c_str())。atoi()需要传入char*,因此string在传入之前还要进行一下转化s.c_str(). No.224 含括号的加减运算(AC) 同上,用两个栈分别存储数字和运算符。 不需要在字符串后面“+0” 当前指向的字符为 ‘)’ 时需要特殊处理。和‘)’ 挨着的一定是 ‘(...