LeetCode[Hard]------418. Sentence Screen Fitting

問題描述

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.

簡單翻譯一下，有一個sentence數組，裏面存若干個String, 現在要求在row行，col列中放這些String, 要求每個String不可分隔，每個String間至少要有一個分隔符(-)，一行放不下的話就放下一行。問整個sentence數組能在row和col中完整出現幾次。

思路：

看完題目首先想到straight force解法，按row進行loop，一個一個往row裏面加，空間不夠就跳下一行。
然而，當row=20000，col=20000是，系統報time exceed錯誤。仔細讀題發現幾個細節：

1. 每行的開頭必然是String，不是分隔符(-).
2. sentence數組大小最大隻有100。
3. 對於每個String，它在該行的產生的結果都是一樣的！

由此，我們另闢蹊徑，把sentence中的每個String作爲開頭，存儲一下在該行sentence重複了多少次，記錄一下下一行的開頭String應該是什麼。有了這些數據，我們最後只需要進行一下簡單的Loop，即可得到最後的result.

代碼：

	//best answer
	public int wordsTyping(String[] sentence, int rows, int cols) {
		int[] nextRowIndex = new int[sentence.length];
        int[] times = new int[sentence.length];
        for(int i=0;i<sentence.length;i++) {
            int curLen = 0;
            int index = i;
            int time = 0;
            while(curLen + sentence[index].length() <= cols) {
                curLen += sentence[index++].length()+1;
                if(index==sentence.length) {
                	index = 0;
                    time ++;
                }
            }
            nextRowIndex[i] = index;
            times[i] = time;
        }
        int res = 0;
        int index = 0;
        for(int i=0; i<rows; i++) {
            res += times[index];
            index = nextRowIndex[index];
        }
        return res;
	}