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.
一刷
题解：我们首先构造一个string, 用空格把word分隔开，并在尾部添加空格, 形成字符串s。(循环数组，所以真实的位置为start%s.length())
然后我们用start来表示下一行的起始位置。如果start指向的位置为空格，那么start--, 因为可以忽略这个空格。
如果不为空格，那么我们至少要保证s[start-1]为空格，否则我们split了一个单词。
Example:
以下为start所指的位置
"abc de f abc de f abc de f ..."
^ ^ ^ ^ ^
0 7 13 18 25
output:
"abc de"
"f abc "
"de f "
"abc de"
class Solution {
public int wordsTyping(String[] sentence, int rows, int cols) {
String s = String.join(" ", sentence) + " ";
int start = 0, len = s.length();
for(int i=0; i<rows; i++){
start += cols;
if(s.charAt(start%len) == ' '){
start++;//start with space, can eliminate
}else{
while(start>0 &&s.charAt((start-1)%len)!=' '){
start--;
}
}
}
return start/len;
}
}
