# 422. Valid Word Square 有效的单词方块

@TOC

## # 题目描述

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",
]

Output:
false

Explanation:

Therefore, it is NOT a valid word square.
``````

## # 解题方法

### # 拼接出每一列的字符串

C++代码如下：

``````class Solution {
public:
bool validWordSquare(vector<string>& words) {
int maxCol = 0;
for (string& word : words) {
if (maxCol < word.size())
maxCol = word.size();
}
int total = maxCol > words.size() ? maxCol : words.size();
for (int col = 0; col < total; ++col) {
string curcol;
for (int row = 0; row < words.size(); ++row) {
if (words[row].size() > col) {
curcol += words[row][col];
}
}
if (curcol != words[col]) {
return false;
}

}
return true;
}
};
``````

## # 日期

2019 年 9 月 19 日 —— 举杯邀明月，对影成三人