LintCode 888: Valid Word Square

888. Valid Word Square
     中文English
     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).

Example
Example1

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.
Example2

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.
Example3

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

解法1：直接比较。
代码如下：

class Solution {
public:/*** @param words: a list of string* @return: a boolean*/bool validWordSquare(vector &words) {int m = words.size();if (m == 0) return true;int n = 0;for (int i = 0; i < m; ++i) {n = max(n, (int)words[i].size());}int i = 0, j = 0;for (i = 0; i < n; ++i) {string columString;for (j = 0; j < m; ++j) {if (words[j].size() >= i) columString += words[j][i];}if (columString != words[i]) return false;  }return true;}
};

本文来自互联网用户投稿，文章观点仅代表作者本人，不代表本站立场，不承担相关法律责任。如若转载，请注明出处。 如若内容造成侵权/违法违规/事实不符，请点击【内容举报】进行投诉反馈！