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.
分析：
统计加空格的句子总长度，然后遍历每一行，初始化colsRemaining为cols，然后还需要一个变量idx，来记录当前单词的位置，如果colsRemaining大于0，就进行while循环，如果当前单词的长度小于等于colsRemaining，说明可以放下该单词，那么就减去该单词的长度就是剩余的空间，然后如果此时colsRemaining仍然大于0，则减去空格的长度1，然后idx自增1，如果idx此时超过单词个数的范围了，说明一整句可以放下，那么就有可能出现宽度远大于句子长度的情况，所以我们加上之前放好的一句之外，还要加上colsRemaining/len的个数，然后colsRemaining%len是剩余的位置，此时idx重置为0.
class Solution {
int wordsTyping(String[] sentence, int rows, int cols) {
String all = "";
for (String word : sentence) {
all += (word + " ");
}
int res = , idx = , n = sentence.length, len = all.length();
for (int i = ; i < rows; ++i) {
int colsRemaining = cols;
while (colsRemaining > ) {
if (sentence[idx].length() <= colsRemaining) {
colsRemaining -= sentence[idx].length();
if (colsRemaining > ) {
colsRemaining -= ;
}
if (++idx >= n) {
res += ( + colsRemaining / len);
colsRemaining %= len;
idx = ;
}
} else {
break;
}
}
}
return res;
}
}