418. Sentence Screen Fitting

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: 1Explanation:hello---world---The character '-' signifies an empty space on the screen.

Example 2:

Input:rows = 3, cols = 6, sentence = ["a", "bcd", "e"]Output: 2Explanation: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: 1Explanation:I-hadapplepie-Ihad--The character '-' signifies an empty space on the screen.

首先，需要一个dp[]记录以每个sentence开头的row可以装多少个词，再进行row次循环，count+=dp[k] - k；k = dp[k] % n。代码如下：

public class Solution {    public int wordsTyping(String[] sentence, int rows, int cols) {        int len = 0, prev = 0, n = sentence.length;        int[] dp = new int[n];        for (int i = 0; i < n; i ++) {            if (i != 0 && len > 0) {                len -= (sentence[i - 1].length() + 1);            }            while (len + sentence[prev % n].length() <= cols) {                len += sentence[prev % n].length() + 1;                prev ++;            }            dp[i] = prev;        }        int count = 0, k = 0;        for (int i = 0; i < rows; i ++) {            count += dp[k] - k;            k = dp[k] % n;        }        return count / n;    }}