zoukankan      html  css  js  c++  java
  • [Leetcode] 58. Length of Last Word

    Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word (last word means the last appearing word if we loop from left to right) in the string.

    If the last word does not exist, return 0.

    Note: A word is defined as a maximal substring consisting of non-space characters only.

    Example:

    Input: "Hello World"
    Output: 5

    最后一个单词的长度。弱智题,直接上代码了。记得要把input先trim一下,去掉前后多余的空格。

    时间O(n)

    空间O(1)

    JavaScript实现

     1 /**
     2  * @param {string} s
     3  * @return {number}
     4  */
     5 var lengthOfLastWord = function (s) {
     6     // corner case
     7     if (s === null || s.length === 0) {
     8         return 0;
     9     }
    10 
    11     // normal case
    12     let input = s.trim();
    13     let count = 0;
    14     for (let i = input.length - 1; i >= 0; i--) {
    15         if (input[i] !== ' ') {
    16             count++;
    17         } else {
    18             break;
    19         }
    20     }
    21     return count;
    22 };

    Java实现

     1 class Solution {
     2     public int lengthOfLastWord(String s) {
     3         // corner case
     4         if (s == null || s.length() == 0) {
     5             return 0;
     6         }
     7         
     8         // normal case
     9         String input = s.trim();
    10         int count = 0;
    11         for (int i = input.length() - 1; i >= 0; i--) {
    12             if (input.charAt(i) != ' ') {
    13                 count++;
    14             }
    15             else {
    16                 break;
    17             }
    18         }
    19         return count;
    20     }
    21 }

    LeetCode 题目总结

  • 相关阅读:
    ES6 Promise 对象及解决回调地狱问题
    ES6 Iterator迭代器和for...of循环
    ES6 Reflect反射
    ES6 Proxy代理
    ES6 Map对象与Set对象
    端口隔离的应用场景与配置
    交换机级联,堆叠,集群技术介绍
    OSPF虚连接简单配置
    小结ospf基本配置的三个参数
    静态路由配置的3个参数
  • 原文地址:https://www.cnblogs.com/cnoodle/p/12016284.html
Copyright © 2011-2022 走看看