LeetCode 418. Sentence Screen Fitting 调整屏幕上的句子_leetcode 屏幕可显示句子-优快云博客

本文详细解析了LeetCode上一道关于如何在指定大小的屏幕上展示句子的问题。通过对句子进行预处理，计算出每行能容纳的单词数量，并通过动态规划的方式存储这些信息，从而快速得出屏幕能完整显示句子的次数。

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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;
}