Sentence Screen Fitting

旧城等待， 2021-11-22 15:58 231阅读 0赞

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.
    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.

1 class Solution {
     2     int wordsTyping(String[] sentence, int rows, int cols) {
     3         String all = "";
     4         for (String word : sentence) {
     5             all += (word + " ");
     6         }
     7         int res = 0, idx = 0, n = sentence.length, len = all.length();
     8         for (int i = 0; i < rows; ++i) {
     9             int colsRemaining = cols;
    10             while (colsRemaining > 0) {
    11                 if (sentence[idx].length() <= colsRemaining) {
    12                     colsRemaining -= sentence[idx].length();
    13                     if (colsRemaining > 0) {
    14                         colsRemaining -= 1;
    15                     }
    16                     if (++idx >= n) {
    17                         res += (1 + colsRemaining / len);
    18                         colsRemaining %= len;
    19                         idx = 0;
    20                     }
    21                 } else {
    22                     break;
    23                 }
    24             }
    25         }
    26         return res;
    27     }
    28 }

转载于:https://www.cnblogs.com/beiyeqingteng/p/11206611.html