LeetCode（String) 2114. Maximum Number of Words Found in Sentences

不念不忘少年蓝@ 2023-09-23 21:44 95阅读 0赞

# 1.问题 #

A sentence is a list of words that are separated by a single space with no leading or trailing spaces.

You are given an array of strings sentences, where each sentences\[i\] represents a single sentence.

Return the maximum number of words that appear in a single sentence.  
Example 1:

> Input: sentences = \[“alice and bob love leetcode”, “i think so too”, “this is great thanks very much”\]  
> Output: 6  
> Explanation:
> 
>  *  The first sentence, “alice and bob love leetcode”, has 5 words in total.
>  *  The second sentence, “i think so too”, has 4 words in total.
>  *  The third sentence, “this is great thanks very much”, has 6 words in total.  
>     Thus, the maximum number of words in a single sentence comes from the third sentence, which has 6 words.

Example 2:

> Input: sentences = \[“please wait”, “continue to fight”, “continue to win”\]  
> Output: 3  
> Explanation: It is possible that multiple sentences contain the same number of words.  
> In this example, the second and third sentences (underlined) have the same number of words.

Constraints:

1 <= sentences.length <= 100  
1 <= sentences\[i\].length <= 100  
sentences\[i\] consists only of lowercase English letters and ’ ’ only.  
sentences\[i\] does not have leading or trailing spaces.  
All the words in sentences\[i\] are separated by a single space.

# 2.解题思路 #

## 方法1 ##

> 1.设置变量sum的初始值为0  
> 2.for遍历数组中的字符串  
> 3.取出字符串，遍历字符串元素；如果字符串元素为空格，在num的基础上加一  
> 4.比较sum和num的大小，最大的值为sum的值  
> 5.返回sum的值

## 方法2 ##

> 1.设置变量sum的初始值为0  
> 2.遍历数组中的字符串  
> 3.取出字符串对字符串内进行分割，分割后的元素个数长度进行比较  
> 4.返回sum的值

# 3.代码 #

## 代码1 ##

class Solution {
        
        public int mostWordsFound(String[] sentences) {
        
            int sum = 0; //1.设置变量sum的初始值为0
            for(int i=0;i<sentences.length;i++){
        //2.for遍历数组中的字符串
                int num=1;//1个空格两个字符串，初始值为1而不是0的理由，可以为0，但是在最后的返回值+1；
                for(char c:sentences[i].toCharArray()){
        //3.取出字符串，遍历字符串元素；如果字符串元素为空格，在num的基础上加一
                    if(c == ' ') {
        
                        num++;
                    }
                }
                sum= Math.max(sum,num);//4.比较sum和num的大小，最大的值为sum的值
            }
            return sum;//5.返回sum的值
            
        }
    }

## 代码2 ##

class Solution {
        
        public int mostWordsFound(String[] sentences) {
        
            int sum=0;//1.设置变量sum的初始值为0
            for(int i=0; i<sentences.length; i++) {
        //2.遍历数组中的字符串
                sum = Math.max(sum,(sentences[i].split(" ")).length);//3.取出字符串对字符串内进行分割，分割后的元素个数长度进行比较
            }
            return sum;//4.返回sum的值
        }
    }

和上面的代码解题思路相同，只是使用了for each的遍历方法

class Solution {
        
        public int mostWordsFound(String[] sentences) {
        
            int sum=0;//1.设置变量sum的初始值为0
            for(String s: sentences) {
        //2.遍历数组中的字符串
                max = Math.max(sum,s.split(" ").length);//3.取出字符串对字符串内进行分割，分割后的元素个数长度进行比较
            }
            return sum;//4.返回sum的值
        }