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418. Sentence Screen Fitting

418. Sentence Screen Fitting

作者: Nancyberry | 来源:发表于2018-05-29 07:15 被阅读0次

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:

  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.

Solution

好迷啊,不太懂,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;
    }
}

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