Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. If the last word does not exist, return 0. Note: A word is defined as a character sequence consists of non-space cha…
[ 问题: ] Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. If the last word does not exist, return 0. 给你一个字符串,设法获取它最后一个单词的长度.假设这个单词不存在,则返回0. [ 分析 : ] A word is defined…
Given a string s consists of upper/lower-case alphabets and empty space characters' ', return the length of last word in the string. If the last word does not exist, return 0. Note: A word is defined as a character sequence consists of non-space char…
1.题目: Given an input string, reverse the string word by word. For example,Given s = "the sky is blue",return "blue is sky the". 运用到了栈的思想.先进后出,这样就逆序了. 时间复杂度,就是整个程序在运行过程中,每个小模块运行的次数之和. 时间复杂度指运行时间需求,空间复杂度是指运行空间需求. public class toChar1{ pu…
[058-Length of Last Word (最后一个单词的长度)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. If the last word does not exist, return…