Leetcode-418. Sentence Screen Fitting

前言：正好碰见Leetcode有一次在线笔试，测试一下，还是挺开心的，全做出来了Rank：98/869。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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Given a rows x cols screen and a sentence represented by a list of 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 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.

这个题目理解很容易，关键是怎么避免超时，我只做了一个trick，感觉不完美，但是最后只有10分钟左右，也没管了。最后也Accepted了，略意外。

public class Solution {
    public int wordsTyping(String[] sentence, int rows, int cols) {
     
	int i = 0; j = 0,k = 0,times = 0;
	while( i < rows && j < cols){
		if(sentence[k].length() > cols) break;
		int remain = cols - j;
		if(remain == sentence[k].length() + 1 || remain == sentence[k].length()){
			k ++;j = 0;i++;
			
		}else if( remain < sentence[k].length()){
			j = 0; i ++;
		}else{
			j = j + sentence[k] + 1; k ++;
		}
		if( k == sentence.length) {k = 0; times ++;}
	}

	return times;
   
    }
}