https://leetcode.com/problems/reverse-integer/ understanding: 最intuitive的办法就是直接把integer化成string,然后变成list。这里如果化成string,会有溢出的问题,比如integer是1534236469,这个数字反过来就是个很大的数,溢出了,必须返回0. 如果是直接用int计算的,那就会自动溢出得到正确结果。这里如果变成list,则效率底下。
What constitutes a word? A sequence of non-space characters constitutes a word. Could the input string contain leading or trailing spaces? Yes. However, your reversed string should not contain leading or trailing spaces. How about multiple spaces between two words? Reduce them to a single space...
Given an input string, reverse the string word by word. For example, Given s = “the sky is blue”, return “blue is sky the”. 思路一: 先休整下给定的字符串,去掉其中的多于一个的空格,就是使所有单词之间的空格都变成一个,然后去掉最后面的空格,给最前面加一个空格。这样每次遇到空格的时候,前...
intmain(intargc,char** argv) { strings ="a b cd"; reverseWords(s); cout<<s; return0; } 其他版本还有C的: #include <iostream> #include <cstdlib> #include <string> #include <algorithm> usingnamespacestd; voidreverseWords(string&s) { constchar* src = s.c_str(); if(src == NULL...
Write a function that takes a string as input and reverse only the vowels of a string. Example 1: Givens = "hello", return "holle". Example 2: Givens = "leetcode", return "leotcede". Note 第一种解法:将字符串转化为字符数组,用一头一尾两个指针向中间夹逼,遇到两个元音字母就进行位置交换...
Can you solve this real interview question? Reverse Words in a String III - Given a string s, reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. Example 1: Input: s = "Let's take
输入:s = "a good example" 输出:"example good a" 解释:如果两个单词间有多余的空格,反转后的字符串需要将单词间的空格减少到仅有一个。 提示: 1 <= s.length <= 104 s 包含英文大小写字母、数字和空格 ' ' s 中至少存在一个 单词 进阶:如果字符串在你使用的编程语言中是一种可变数据类型,请尝试...
Write a function that takes a string as input and reverse only the vowels of a string. Example 1: Given s = "hello", return "holle". Example 2: Given s = "leetcode", return "leotcede". Note: The vowels does not include the letter "y". ...
Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". */ #include <stdio.h> #include <string.h> #include <stdlib.h> struct node { char *dest; struct node *Next; ...
Problem: 复杂度 时间复杂度: O(n) 空间复杂度: O(n) Code Python3 Java C++ 数学 2+ 1 270 0LSY ・ 2024.12.31 150. 逆波兰表达式求值 Problem: 思路 当遇到数字时直接进栈,不需要执行任何操作,若是运算符号,则取出栈顶的两个数组进行运算并将结果重新加入到栈中,直到tokens数组遍历完,栈中最后只存...