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:
- 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 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.
这道题给我们了一个句子，由若干个单词组成，然后给我们了一个空白屏幕区域，让我们填充单词，前提是单词和单词之间需要一个空格隔开，而且单词不能断开，如果当前行剩余位置放下不下某个单词，则必须将该单词整个移动到下一行。我刚开始想的是便利句子，每个单词分别处理，但是这种做法很不高效，因为有可能屏幕的宽度特别大，而单词可能就一两个，那么我们这样遍历的话就太浪费时间了，应该直接用宽度除以句子加上空格的长度之和，可以快速的得到能装下的个数。下面这种方法设计的很巧妙，思路是用start变量来记录下能装下的句子的总长度，最后除以一个句子的长度，就可以得到个数。而句子的总长度的求法时要在每个单词后面加上一个空格(包括最后一个单词)，我们遍历屏幕的每一行，然后每次start都加上宽度，然后看all[start%len]是否为空格，是的话就start加1，这样做的好处是可以处理末尾是没有空格的情况，比如宽度为1，只有一个单词a，那么我们都知道是这样放的 a ，start变为1，len是2，all[start%len]是空格，所以start自增1，变成2，这样我们用start/len就知道能放下几个了。对于all[start%len]不为空格的情况，如果all[(start-1)%len]也不为空格，那么start就自减1，进行while循环，直至其为空格为止
public class WordsTyping {
public int wordsTyping(List<String> setence, int rows, int cols) {
String all = "";
for (String word : setence) {
all += (word + " ");
}
int start = 0, len = all.length();
for (int i = 0; i < rows; i++) {
start += cols;
if (all.charAt(start % len) == ' ') {
start++;
} else {
while (start > 0 && all.charAt((start - 1) % len) != ' ') {
start--;
}
}
}
return start / len;
}
}