python-leetcode题解之第418题屏幕可显示句子的数量.zip

''' Given a rows x cols screen 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: 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 is greater than 0 and 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. ''' class Solution(object): def wordsTyping(self, sentences, rows, cols): """" :sentences List<String> :rows int :cols int """ sentence = '-'.join(sentences) sentence += '-' index_in_sentence = 0 for row in range(rows): index_in_sentence += cols if sentence[(index_in_sentence%len(sentence))] == '-': index_in_sentence += 1 else: while index_in_sentence > 0 and sentence[((index_in_sentence - 1)%len(sentence))] != '-': index_in_sentence -= 1 return index_in_sentence/len(sentence) solution = Solution() row, col = 3, 6 sentences = ["a", "bcd", "e"] print solution.wordsTyping(sentences=sentences, rows=row, cols=col)