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
假设string是无限循环下去的，需要记录所有row走完会move到哪里。用start指针的值记录一共在screen上有多少字符。最后通过start / sentence length即可知道一共能打印下多少个sentence。
具体包含以下步骤：
- 先把sentence连起来，然后摊平。例如:[“abc”, “de”, “f”]，连成sentence是”abc de f “，设想这个sentence无限循环，就是”abc de f abc de f abc de f abc…”
- 用start pointer记录每一行的start，每多一个row，把pointer往后移动cols个。
- 移动pointer时两种情况需要处理一下：
- 一行的开头刚好是space，start指针直接+1，即remove一个space
- 一行的开头刚好是一个word的中间，start需要退回到该word的开头，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;
}
}
