【多次过】Lintcode 888:Valid Word Square

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

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

 Notice

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

Given
[
  "abcd",
  "bnrt",
  "crmy",
  "dtye"
]
return 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.

Given
[
  "abcd",
  "bnrt",
  "crm",
  "dt"
]
return 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.

Given
[
  "ball",
  "area",
  "read",
  "lady"
]
return false

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

Therefore, it is NOT a valid word square.

解题思路：

    遍历二维数组，检测行列反转前后两字母是否一致即可。

    注意：行列两指向col，row均不能越彼此的界，否则为false。

class Solution {
public:
    /**
     * @param words: a list of string
     * @return: a boolean
     */
    bool validWordSquare(vector<string> &words)
    {
        // Write your code here
        for(int col=0;col<words.size();col++)
            for(int row=0;row<words[col].size();row++)
                if(col >= words[col].size() || row >= words.size() || words[row][col] != words[col][row])
                return false;

        return true;
    }
};