原题: 434. Number of Segments in a String 解题: 刚看到题目时,觉得可以通过统计空格个数,但想想有可能会有多个空格的情况 思路: 一:遍历字符,if条件碰到非空格时,计数加1,然后while循环跳过非空格字符,一直到最后 二:设置flag初始为0,当碰到非空格时,计数加1,且flag置1,当flag为1且碰到空格时,flag再重置为0 思路一是自己想的,思路二更巧妙,是论坛里摘抄的 AC代码: 思路一: class Solution { public: in…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" Output:…

［抄题］: Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" O…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printablecharacters. Example: Input: "Hello, my name is John" Output:…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 统计 正则表达式 字符串分割 日期 题目地址:https://leetcode.com/problems/number-of-segments-in-a-string/#/description 题目描述 Count the number of segments in a string, where a segment is defined to b…

统计字符串中的单词个数,这里的单词指的是连续的非空字符.请注意,你可以假定字符串里不包括任何不可打印的字符.示例:输入: "Hello, my name is John"输出: 5 详见:https://leetcode.com/problems/number-of-segments-in-a-string/description/ C++: 方法一: class Solution { public: int countSegments(string s) { int cnt=0; f…

434. Number of Segments in a String Easy Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Inp…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" Output:…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" Output:…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. For example, Input: "Hello, my name is John" Output: 5 public int countSegments(String s) { String trimmed = s.trim(); if (…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" Output:…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" Output:…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3941 访问. 统计字符串中的单词个数,这里的单词指的是连续的不是空格的字符. 请注意,你可以假定字符串里不包括任何不可打印的字符. 输入: "Hello, my name is John" 输出: 5 Count the number of segments in a string, where a segment is defined to be a…

这是悦乐书的第226次更新,第239篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第93题(顺位题号是434).计算字符串中的段数,其中段定义为非空格字符的连续序列.请注意,该字符串不包含任何不可打印的字符.例如: 输入:"Hello, my name is John" 输出:5 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 第一种解法 使用正则表达式,将其拆分成字符串数组…

以空格为分隔符,判断一个string可以被分成几部分. 注意几种情况:(1)全都是空格 (2)空字符串(3)结尾有空格 思路: 只要统计出单词的数量即可.那么我们的做法是遍历字符串,遇到空格直接跳过,如果不是空格,则计数器加1,然后用个while循环找到下一个空格的位置,这样就遍历完了一个单词,再重复上面的操作直至结束,就能得到正确结果: class Solution { public: int countSegments(string s) { , n = s.size(); ; i < n;…

1.题目描述 2.题目分析 找到字符串中的空格即可 3.代码 int countSegments(string s) { ){ ; } vector<string> v; ; i < s.size(); i++){ if( isspace(s[i]) ){ continue; } ; while( !isspace(s[j]) ){ if( j < s.size() ) j++; else break; } string sb = s.substr(i,j-i); v.push_b…

统计字符串中的单词个数,这里的单词指的是连续的不是空格的字符. 请注意,你可以假定字符串里不包括任何不可打印的字符. 示例: 输入: "Hello, my name is John" 输出: 5 class Solution { public: int countSegments(string s) { int len = s.size(); string temp = ""; int res = 0; for(int i = 0; i < len; i++)…

Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters. Please note that the string does not contain any non-printable characters. Example: Input: "Hello, my name is John" Output:…

806. Number of Lines To Write String 整体思路: 先得到一个res = {a : 80 , b : 10, c : 20.....的key-value对象}(目的是当调用res["a"]时得到一个等于10的值): 遍历传入的字符串,把每个元素带入到res中,并把所有的值进行累加:得到一个累加值sum,如:sum= sum + res["a"] + res["b"] + res["c"]...…

problem 806. Number of Lines To Write String solution: class Solution { public: vector<int> numberOfLines(vector<int>& widths, string S) { , len = ; for(auto ch:S) { int t = widths[ch-'a']; ) line++; len = (len+t>) ? t : len+t; } return…

一.报错截图 [Vue warn]: Invalid prop: type check failed for prop "jingzinum". Expected Number with value NaN, got String with value "fuNum". 二.报错代码 <!DOCTYPE html> <html lang="en"> <head> <meta charset="U…

Question 806. Number of Lines To Write String Solution 思路:注意一点,如果a长度为4,当前行已经用了98个单元,要另起一行. Java实现: public int[] numberOfLines(int[] widths, String S) { int left = 0; int lines = 0; for (char c : S.toCharArray()) { left += widths[c - 'a']; if (left >=…

这是悦乐书的第319次更新,第340篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第188题(顺位题号是806).我们要将给定字符串S的字母从左到右写成行.每行最大宽度为100个单位,如果写一个字母会导致该行的宽度超过100个单位,则会写入下一行.给出一个数组宽度,一个数组,其中widths[0]是'a'的宽度,widths[1]是'b'的宽度,widths[25]是'z'的宽度. 现在回答两个问题:S中至少有一个字符有多少行,最后一行使用的宽度是多少?将答案作为长…

We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an arra…

We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an arra…

题目要求 We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an…

We are to write the letters of a given string S, from left to right into lines. Each line has maximum width 100 units, and if writing a letter would cause the width of the line to exceed 100 units, it is written on the next line. We are given an arra…

1.题目描述 2.分析 使用一个map将字母和数字对应起来,方便后续使用. 3.代码 vector<int> numberOfLines(vector<int>& widths, string S) { map<char,int> m; vector<int> ans; ; i< ;i++) m[i+'a'] = widths[i]; ; ; ; ; t < S.size(); t++) { curLen += m[ S[t] ]; la…

1 小结: 1)int-->NSNumber:numberWithInt 2)NSNumber-->nsinteger:integerValue 3)string -->double:initWithString 4)CGFloat --> dobule:initWithFloat,decimalobj doubleValue 5)使用NSInteger,因为这样就不用考虑设备是32位的还是64位的. 6)NSInteger是基础类型,但是NSNumber是一个类.如果想要在NSM…

解答 class Solution { public: vector<int> numberOfLines(vector<int>& widths, string S) { vector<int> result; int line=1,units=0; map<char,int> pairs; char c='a'; for(int width:widths){ pairs.insert(make_pair(c,width)); ++c; } for…