[LeetCode] Valid Word Square 验证单词平方

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.

这道题给了我们一个二位数组，每行每列都是一个单词，需要满足第k行的单词和第k列的单词要相等，这里不要求每一个单词的长度都一样，只要对应位置的单词一样即可。那么这里实际上也就是一个遍历二维数组，然后验证对应位上的字符是否相等的问题，由于各行的单词长度不一定相等，所以我们在找对应位置的字符时，要先判断是否越界，即对应位置是否有字符存在，遇到不符合要求的地方直接返回false，全部遍历结束后返回true，参见代码如下：

class Solution {
public:
    bool validWordSquare(vector<string>& words) {
        if (words.empty()) return true;
        if (words.size() != words[0].size()) return false;
        for (int i = 0; i < words.size(); ++i) {
            for (int j = 0; j < words[i].size(); ++j) {
                if (j >= words.size() || i >= words[j].size() || words[i][j] != words[j][i]) {
                    return false;
                }
            }
        }
        return true;
    }
};

本文转自博客园Grandyang的博客，原文链接：验证单词平方[LeetCode] Valid Word Square ，如需转载请自行联系原博主。