算法提高 最长单词 时间限制:1.0s 内存限制:512.0MB 编写一个函数,输入一行字符,将此字符串中最长的单词输出. 输入仅一行,多个单词,每个单词间用一个空格隔开.单词仅由小写字母组成.所有单词的长度和不超过100000.如有多个最长单词,输出最先出现的. 样例输入 I am a student 样例输出 student #include<stdio.h> #include<string.h> #define max 100000 int is_zim
Given a list of strings words representing an English Dictionary, find the longest word in words that can be built one character at a time by other words in words. If there is more than one possible answer, return the longest word with the smallest l
Given a string and a string dictionary, find the longest string in the dictionary that can be formed by deleting some characters of the given string. If there are more than one possible results, return the longest word with the smallest lexicographic
题目描述: 分析:先建一个数组s用来存储每个字符串的长度,然后遍历数组s得到最大的数max,这个数就是词典中的最长单词的长度,由于可能有多个长度相等的单词,所以要循环整个词典,当一个单词的长度等于max时,就将它存到要返回的ArrayList中. 我的代码: public class Solution { /* * @param dictionary: an array of strings * @return: an arraylist of strings */ public ArrayLi
Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0. Example
Swift3.0语言教程获取字符串长度 Swift3.0语言教程获取字符串长度,当在一个字符串中存在很多的字符时,如果想要计算字符串的长度时相当麻烦的一件事情,在NSString中可以使用length属性去实现这一功能,其语法形式如下: var length: Int { get } [示例1-15]以下将实现字符串长度的计算. import Foundation var str=NSString(stringLiteral: "Swift Hello")