418. Sentence Screen Fitting

418. Sentence Screen Fitting

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:

1. A word cannot be split into two lines.
2. The order of words in the sentence must remain unchanged.
3. Two consecutive words in a line must be separated by a single space.
4. Total words in the sentence won't exceed 100.
5. Length of each word is greater than 0 and won't exceed 10.
6. 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.

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假如给定的string s = “a_bc“， 在有限的行数下我们要构建一个最终的string为 “a_bc_a_bc_a_bc_a_bc_…”，然后除以”a_bc_”将能得到最终的答案 代码里s.charAt[len % l] 返回的是一位字符，[len - 1 % 1]返回的是当前(末尾)字符 if里面是为了补齐空格

class Solution {
    public int wordsTyping(String[] sentence, int rows, int cols) {
        String s = String.join(" ", sentence) + " ";
        int l = s.length();
        int len = 0;
        
        for (int i = 0; i < rows; i++) {
            len += cols;
            if (s.charAt(len % l) == ' ') {
                len++;
            }
            while (len > 0 && s.charAt((len - 1) % l) != ' ') {
                len--;
            }
            
        }
        
        return len / l;
    }
}