05 August 2008
给定一个 row x cols 屏幕和一个表示为字符串列表的句子，返回给定的句子可以在屏幕上出现的次数。
句子中单词的顺序必须保持不变，一个单词不能分成两行。一个空格必须分隔一行中的两个连续单词。
Example 1:
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Input: sentence = ["hello","world"], rows = 2, cols = 8
Output: 1
Explanation:
hello---
world---
The character '-' signifies an empty space on the screen.
Example 2:
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Input: sentence = ["a", "bcd", "e"], rows = 3, cols = 6
Output: 2
Explanation:
a-bcd-
e-a---
bcd-e-
The character '-' signifies an empty space on the screen.
这道题给了一个句子，由若干个单词组成，然后给我们了一个空白屏幕区域，让我们填充单词，前提是单词和单词之间需要一个空格隔开，而且单词不能断开，如果当前行剩余位置放不下某个单词，则必须将该单词整个移动到下一行。
DP
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class Solution {
static class CountState {
int index;
int count;
CountState(int index, int count) {
this.index = index;
this.count = count;
}
}
CountState countWords(int[] wordLength, int start, int cols, int idx) {
int count = 0;
for (int j = start; j + wordLength[idx % wordLength.length] <= cols; ) {
j += wordLength[idx % wordLength.length] + 1;
count = (++idx % wordLength.length == 0) ? count + 1 : count;
}
return new CountState(idx, count);
}
public int wordsTyping(String[] sentence, int rows, int cols) {
int index = 0;
int characters = 0;
int[] wordLength = new int[sentence.length];
for (int i = 0; i < sentence.length; ++i) {
wordLength[i] = sentence[i].length();
characters += wordLength[i] + 1;
}
int sentenceCount = cols / characters;
int start = characters * sentenceCount;
int count = sentenceCount * rows;
CountState[] dp = new CountState[sentence.length];
for (int i = 0; i < rows; ++i) {
int realIndex = index % sentence.length;
if (dp[realIndex] == null) {
dp[realIndex] = countWords(wordLength, start, cols, realIndex);
}
index = dp[realIndex].index;
count += dp[realIndex].count;
}
return count;
}
}
https://leetcode.com/problems/sentence-screen-fitting/discuss/90845/21ms-18-lines-Java-solution
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class Solution {
public int wordsTyping(String[] sentence, int rows, int cols) {
String s = String.join(" ", sentence) + " ";
int[] offset = new int[s.length()];
IntStream.range(1, s.length()).forEach(i -> offset[i] = s.charAt(i) == ' ' ? 1 : offset[i - 1] - 1);
return IntStream.range(0, rows).reduce(0, (a, b) -> a + cols + offset[(a + cols) % s.length()]) / s.length();
}
}