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sentenceScreenFitting.cpp
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sentenceScreenFitting.cpp
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//
// OJ#418.cpp
// LeeteCodeOJ#418
//
// Created by Wang Yi on 25/10/16.
// Copyright (c) 2016 Wang Yi. All rights reserved.
//
/**********************************************************************************
*
* 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.
*
* *********************************************************************************/
#include "sentenceScreenFitting.h"
#include <string.h>
#include <iomanip>
#include <iostream>
#define FOR(START, END) for((START)=0;(START)<(END);(START)+=1) {
#define END ;}
#define N 100
// first brute force solution
int SentenceScreenFitting(char stc[][M], int row, int col, size_t l)
{
int i, j, t, s, k=0, ret=0;
int len[N] = {0};
FOR(i, l) len[i] = (int)strlen(stc[i]) END
FOR(i, row)
j = col;
t = k;
while (j - len[k] >= 0) {
j -= len[k] + 1;
if (k+1 > l-1){
FOR(s, k-t) std::cout << stc[t+s] << ' ' END;
std::cout << stc[t+s];
if (j - len[0] >= 0)
std::cout << ' ';
k = 0;
t = 0;
ret++;
} else{
k++;
}
}
FOR(s, k-t-1) std::cout << stc[t+s] << ' ' END;
if (k-t > 0)
std::cout << stc[t+s];
std::cout << std::setfill ('-') << std::setw(j == col ? col:j+1) << "" << std::endl;
END
return ret;
}
/*
int Solution::wordsTyping(vector<string>& sentence, int rows, int cols)
{
int i, j, t, s, k=0, ret=0, l=(int)sentence.size();
int len[N] = {0};
FOR(i, l) len[i] = (int)sentence[i].size() END
FOR(i, rows)
j = cols;
t = k;
while (j - len[k] >= 0) {
j -= len[k] + 1;
if (k+1 > l-1){
// FOR(s, k-t) std::cout << sentence[t+s] << ' ' END;
// std::cout << sentence[t+s];
// if (j - len[0] >= 0)
// std::cout << ' ';
k = 0;
t = 0;
ret++;
} else{
k++;
}
}
// FOR(s, k-t-1) std::cout << sentence[t+s] << ' ' END;
// if (k-t > 0)
// std::cout << sentence[t+s];
//
// std::cout << std::setfill ('-') << std::setw(j == cols ? cols:j+1) << "" << std::endl;
END
return ret;
}
*/
// second boosting
int Solution::wordsTyping(vector<string>& sentence, int rows, int cols)
{
int i, j, k=0, ret=0, l=(int)sentence.size();
// int t, s;
int jump = 0;
int len[N] = {0};
FOR(i, l) len[i] = (int)sentence[i].size();jump += len[i] END
jump += l;
FOR(i, rows)
j = cols;
// t = k;
if (k == 0){
ret += j / jump;
j = j % jump;
}
while (j - len[k] >= 0) {
j -= len[k] + 1;
if (k+1 > l-1){
// FOR(s, k-t) std::cout << sentence[t+s] << ' ' END;
// std::cout << sentence[t+s];
// if (j - len[0] >= 0)
// std::cout << ' ';
k = 0;
// t = 0;
ret += j / jump + 1;
j = j % jump;
} else {
k++;
}
}
// FOR(s, k-t-1) std::cout << sentence[t+s] << ' ' END;
// if (k-t > 0)
// std::cout << sentence[t+s];
//
// std::cout << std::setfill ('-') << std::setw(j == cols ? cols:j+1) << "" << std::endl;
END
return ret;
}