[Leetcode] 418. Sentence Screen Fitting 解题报告_leetcode 418视频-优快云博客

本文介绍了一种巧妙的屏幕适配算法，该算法用于计算给定的句子能够在特定大小的屏幕上显示多少次。通过将所有单词连接成一个字符串，并检查字符串在每一行结束时的位置来实现这一目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Given a rows x cols screen and a sentence represented by a list of non-empty words, find how many times the given sentence can be fitted on the screen.

Note:

A word cannot be split into two lines.
The order of words in the sentence must remain unchanged.
Two consecutive words in a line must be separated by a single space.
Total words in the sentence won't exceed 100.
Length of each word is greater than 0 and won't exceed 10.
1 ≤ rows, cols ≤ 20,000.

Example 1:

Input:
rows = 2, cols = 8, sentence = ["hello", "world"]

Output: 
1

Explanation:
hello---
world---

The character '-' signifies an empty space on the screen.

Example 2:

Input:
rows = 3, cols = 6, sentence = ["a", "bcd", "e"]

Output: 
2

Explanation:
a-bcd- 
e-a---
bcd-e-

The character '-' signifies an empty space on the screen.

Example 3:

Input:
rows = 4, cols = 5, sentence = ["I", "had", "apple", "pie"]

Output: 
1

Explanation:
I-had
apple
pie-I
had--

The character '-' signifies an empty space on the screen.

思路：

在网上看到一个巧妙的方法：把所有的字符串都加起来，然后每次看如果位移一整行的距离是否正好落在这个字符串的空格位置，如果不是的话就退后，直到遇到一个空格。

代码：

class Solution {
public:
    int wordsTyping(vector<string>& sentence, int rows, int cols) {
        string str;  
        for(auto val: sentence) {
            str += val + " ";  
        }
        int len = str.size(), start = 0;  
        for(int i = 0; i < rows; i++) {  
            start += cols;  
            if(str[start % len] == ' ') {
                ++start;  
                continue;  
            }  
            while(start > 0 && str[(start - 1) % len] != ' ') {
                --start; 
            }
        }  
        return start / len;
    }
};