LeetCode 418. Sentence Screen Fitting 调整屏幕上的句子
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[LeetCode] Sentence Screen Fitting 调整屏幕上的句子
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.
思路:
- 自己想的时候,先想到用brute force的方法,一行一行的去放,这样做太简单太straitforward,而且还比较慢。所以,也想到了应该有优化的方法,比如,对句子先做预处理,预先计算出包含第1个单词到第i个单词的sub sentence的总长度,然后根据每一行的长度来确定每一行的单词的个数,但是这个方法不够简洁。最后,参考了方法https://discuss.leetcode.com/topic/62364/java-optimized-solution-17ms/2
- 想想什么信息需要记录,以方便解题?这个题,句子长度最多100个单词,但是行、列长度多达20,000。所以,最好对句子做预处理,首先得到以每个单词开头的一行的单词数量,以及下一行开头的单词index,然后遍历每一行,把每一行的单词总数累加,然后把单词总数除以句子单词数得到句子重复的次数。
- 这题有点dp的味道,先枚举每行所有可能的情况,然后存起来,然后查询的方法!!
int wordsTyping(vector<string>& sentence, int rows, int cols) { int n=sentence.size(); vector<int> dp(n); for(int i=0;i<n;i++){ int curWord=i; int curLen=0; int curNum=0; while(curLen+sentence[curWord].size()<=cols) { curNum++; curLen=curLen+sentence[curWord].size()+1; curWord++; if(curWord==n) curWord=0; } dp[i]=curNum; } int res=0; int start=0; for(int i=0;i<rows;i++){ res+=dp[start]; start=(start+dp[start])%n; } return res/n;}
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