422. Valid Word Square
Дан array строк words, return true, если он образует правильный квадрат слов.
Последовательность строк образует правильный квадрат слов, если k-я stringa и k-й столбец читаются одинаково, где 0 <= k < max(numRows, numColumns).
Esempio:
Input: words = ["abcd","bnrt","crmy","dtye"]
Output: true
Explanation:
The 1st row and 1st column both read "abcd".
The 2nd row and 2nd column both read "bnrt".
The 3rd row and 3rd column both read "crmy".
The 4th row and 4th column both read "dtye".
Therefore, it is a valid word square.
C# soluzione
abbinato/originaleusing System;
using System.Collections.Generic;
public class Solution {
public bool ValidWordSquare(IList<string> words) {
int cols = 0;
int rows = words.Count;
var newWords = new List<string>();
foreach (var word in words) {
cols = Math.Max(cols, word.Length);
}
if (cols != words[0].Length || rows != cols) {
return false;
}
for (int col = 0; col < cols; ++col) {
var newWord = string.Empty;
for (int row = 0; row < rows; ++row) {
if (col < words[row].Length) {
newWord += words[row][col];
}
}
newWords.Add(newWord);
}
return words.SequenceEqual(newWords);
}
}
C++ soluzione
bozza automatica, rivedere prima dell'invio#include <bits/stdc++.h>
using namespace std;
// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
public bool ValidWordSquare(vector<string> words) {
int cols = 0;
int rows = words.size();
var newWords = new List<string>();
foreach (var word in words) {
cols = max(cols, word.size());
}
if (cols != words[0].size() || rows != cols) {
return false;
}
for (int col = 0; col < cols; ++col) {
var newWord = string.Empty;
for (int row = 0; row < rows; ++row) {
if (col < words[row].size()) {
newWord += words[row][col];
}
}
newWords.push_back(newWord);
}
return words.SequenceEqual(newWords);
}
}
Java soluzione
abbinato/originaleimport java.util.*;
public class Solution {
public boolean validWordSquare(List<String> words) {
int cols = 0;
int rows = words.size();
List<String> newWords = new ArrayList<>();
for (String word : words) {
cols = Math.max(cols, word.length());
}
if (cols != words.get(0).length() || rows != cols) {
return false;
}
for (int col = 0; col < cols; ++col) {
StringBuilder newWord = new StringBuilder();
for (int row = 0; row < rows; ++row) {
if (col < words.get(row).length()) {
newWord.append(words.get(row).charAt(col));
}
}
newWords.add(newWord.toString());
}
return words.equals(newWords);
}
}
Python soluzione
abbinato/originaleclass Solution:
def validWordSquare(self, words: list[str]) -> bool:
cols = 0
rows = len(words)
newWords = []
for word in words:
cols = max(cols, len(word))
if cols != len(words[0]) or rows != cols:
return False
for col in range(cols):
newWord = ""
for row in range(rows):
if col < len(words[row]):
newWord += words[row][col]
newWords.append(newWord)
return words == newWords
Go soluzione
abbinato/originalepackage main
func validWordSquare(words []string) bool {
cols := 0
rows := len(words)
newWords := []string{}
for _, word := range words {
if len(word) > cols {
cols = len(word)
}
}
if cols != len(words[0]) || rows != cols {
return false
}
for col := 0; col < cols; col++ {
newWord := ""
for row := 0; row < rows; row++ {
if col < len(words[row]) {
newWord += string(words[row][col])
}
}
newWords = append(newWords, newWord)
}
for i := range words {
if words[i] != newWords[i] {
return false
}
}
return true
}
Algorithm
Инициализируйте переменные: cols для максимальной длины слов в arrayе, rows для количества строк в arrayе words, и пустой array newWords для хранения новых слов, представленных каждым столбцом.
Итерация по arrayу words, Definizione максимальной длины слова для cols, проверка, что количество строк равно количеству столбцов. Если Testo не выполняется, возвращаем false.
Для каждого столбца col от 0 до cols - 1, формируем строку newWord из символов на позиции (row, col) для каждой строки. Сохраняем newWord в arrayе newWords. В конце, если newWords и words равны, возвращаем true, иначе false.
😎
Vacancies for this task
offerte attive with overlapping task tags are mostrati.