[leetcode] 418. Sentence Screen Fitting 解题报告

本文链接：https://blog.youkuaiyun.com/qq508618087/article/details/52812711

这是一篇关于LeetCode第418题《Sentence Screen Fitting》的解题报告。题目要求在给定屏幕大小的情况下，计算能完整显示句子的次数。限制条件包括单词顺序不变、单词不能拆分等。文章介绍了两种解题思路，包括直接扫描和滑动窗口方法，并给出了代码实现。

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题目链接: https://leetcode.com/problems/sentence-screen-fitting/

Given a rows x cols screen and a sentence represented by a list of words, findhow 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：一道google的面试题．直接的方法是每次扫描一行，尝试能放几个，这样时间复杂度会高一点．另外一种方法是把所有的字符串都加起来，然后每次看如果位移一整行的距离是否正好落在这个字符串的空格位置，如果不是的话就退后，直到遇到一个空格．

code:

class Solution {
public:
    int wordsTyping(vector<string>& sentence, int rows, int cols) {
        string str;
        for(auto val: sentence) str += val + " ";
        int len = str.size(), start = 0;
        for(int i = 0; i < rows; i++)
        {
            start += cols;
            if(str[start%len]==' ')
            {
                start++;
                continue;
            }
            while(start > 0 && str[(start-1)%len]!=' ') start--;
        }
        return start/len;
    }
};