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.
   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
假設(shè)string是無(wú)限循環(huán)下去的，需要記錄所有row走完會(huì)move到哪里。用start指針的值記錄一共在screen上有多少字符。最后通過start / sentence length即可知道一共能打印下多少個(gè)sentence。

具體包含以下步驟：

1. 先把sentence連起來(lái)，然后攤平。例如:[“abc”, “de”, “f”]，連成sentence是”abc de f “，設(shè)想這個(gè)sentence無(wú)限循環(huán)，就是”abc de f abc de f abc de f abc…”
2. 用start pointer記錄每一行的start，每多一個(gè)row，把pointer往后移動(dòng)cols個(gè)。
3. 移動(dòng)pointer時(shí)兩種情況需要處理一下：
   • 一行的開頭剛好是space，start指針直接+1，即remove一個(gè)space
   • 一行的開頭剛好是一個(gè)word的中間，start需要退回到該word的開頭，即在上一行增加一些space。

class Solution {
    public int wordsTyping(String[] sentence, int rows, int cols) {
        String allSentence = "";
            
        //1. reorganized the sentence using " "
        for (String word : sentence) {
            allSentence += word + " ";
        }
        int len = allSentence.length();
        
        //2. calculate the start pointer
        int start = 0;
        
        for (int i = 0; i < rows; i++) {
            start = start + cols;
            if (allSentence.charAt(start % len) == ' ') {
                start++;
            } else {
                while (start > 0 && allSentence.charAt((start - 1) % len) != ' ') {
                    start--;
                }
            }
        }
        
        return start / len;
    }
}