Description
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.
Solution
好迷啊耗溜，不太懂肘交，to be added...
class Solution {
public int wordsTyping(String[] words, int rows, int cols) {
String sentence = String.join(" ", words) + " "; // add a space between two sentences
int len = sentence.length();
// think there are a lot of same sentences connected
int start = 0; // current position in the sentences group
for (int i = 0; i < rows; ++i) {
start += cols; // start now point to the fist char in the new row
if (sentence.charAt(start % len) == ' ') { // remove space in the beginning of the row
++start;
} else { // if the beginning is not a space, it must be a broken word
// move the broken word from prev line to the new line
while (start > 0 && sentence.charAt((start - 1) % len) != ' ') {
--start;
}
// start will get back to 0 if there's a long word won't fit screen
}
}
return start / len;
}
}
