Leetcode
Dynamic Programming
418 Sentence Screen Fitting

1. Question

Given arows x colsscreen 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. 1.
    A word cannot be split into two lines.
  2. 2.
    The order of words in the sentence must remain unchanged.
  3. 3.
    Two consecutive words in a line must be separated by a single space.
  4. 4.
    Total words in the sentence won't exceed 100.
  5. 5.
    Length of each word is greater than 0 and won't exceed 10.
  6. 6.
    1 ≤ rows, cols ≤ 20,000.
Example 1:
1
Input: rows = 2, cols = 8, sentence = ["hello", "world"]
2
3
Output: 1
4
5
Explanation:
6
hello---
7
world---
8
9
The character '-' signifies an empty space on the screen.
Copied!
Example 2:
1
Input: rows = 3, cols = 6, sentence = ["a", "bcd", "e"]
2
3
Output: 2
4
5
6
Explanation:
7
a-bcd-
8
e-a---
9
bcd-e-
10
11
The character '-' signifies an empty space on the screen.
Copied!
Example 3:
1
Input: rows = 4, cols = 5, sentence = ["I", "had", "apple", "pie"]
2
3
Output: 1
4
5
6
Explanation:
7
I-had
8
apple
9
pie-I
10
had--
11
12
The character '-' signifies an empty space on the screen.
Copied!

2. Implementation

思路: 这道题是给定一个每个单词以空格隔开的句子,要求用多少这个句子才能完全fit上 row * col的screen, 一个单词不能分布在不同的两行。所以我们先得到这个句子用空格隔开后的长度len,然后用start这个变量判断每行最后一个空格在哪,最后的结果就是start / len
1
class Solution {
2
public int wordsTyping(String[] sentence, int rows, int cols) {
3
String s = "";
4
5
for (String word : sentence) {
6
s += word + " ";
7
}
8
9
int len = s.length();
10
int start = 0;
11
12
for (int i = 0; i < rows; i++) {
13
start += cols;
14
if (s.charAt(start % len) == ' ') {
15
++start;
16
}
17
else {
18
while (start > 0 && s.charAt((start - 1) % len) != ' ') {
19
--start;
20
}
21
}
22
}
23
return start / len;
24
}
25
}
Copied!

3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(n)