[LeetCode] 422. Valid Word Square_Easy

Given a sequence of words, check whether it forms a valid word square.

A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns).

Note:

1. The number of words given is at least 1 and does not exceed 500.
2. Word length will be at least 1 and does not exceed 500.
3. Each word contains only lowercase English alphabet a-z.

Example 1:

Input:
[
  "abcd",
  "bnrt",
  "crmy",
  "dtye"
]

Output:
true

Explanation:
The first row and first column both read "abcd".
The second row and second column both read "bnrt".
The third row and third column both read "crmy".
The fourth row and fourth column both read "dtye".

Therefore, it is a valid word square.

Example 2:

Input:
[
  "abcd",
  "bnrt",
  "crm",
  "dt"
]

Output:
true

Explanation:
The first row and first column both read "abcd".
The second row and second column both read "bnrt".
The third row and third column both read "crm".
The fourth row and fourth column both read "dt".

Therefore, it is a valid word square.

Example 3:

Input:
[
  "ball",
  "area",
  "read",
  "lady"
]

Output:
false

Explanation:
The third row reads "read" while the third column reads "lead".

Therefore, it is NOT a valid word square.

思路是有可能words里面的word长度不一致, 就无法正常比较,比如 words = ["abcd","bnrt","crmy","de"] , 那么 temp = list(map("".join, zip(*words))) 只等于 ['abcd', 'bnre'], 所以要将words里面不足长度的用" " 来补齐.

Code

class Solution:
    def validWordSquare(self, words):
        n = len(words)
        for index, word in enumerate(words):
            if len(word) > n:
                return False
            if len(word) < n:
                words[index] += ' '*(n - len(word))
        temp = list(map("".join, zip(*words)))
        return temp == words