原题链接在这里：https://leetcode.com/problems/valid-word-square/
题目：
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:
- 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.
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.
题解：
检查是否关于对角线对称.
每一个char 都要检查一遍. 这里注意内层loop j 不是从 i 开始而是从0开始.
原本想只检查右上部分在左下部分是否有对称即可，但忽略了这里不一定size是对称的，如果左下有这个char而右上没有这个char就不能检测出false.
Time Complexity: O(m * n). m = words.size(). n 是最长string的length.
Space: O(1).
AC Java:
1 public class Solution { 2 public boolean validWordSquare(List<String> words) { 3 if(words == null || words.size() == 0){ 4 return true; 5 } 6 int m = words.size(); 7 for(int i = 0; i<m; i++){ 8 for(int j = 0; j<words.get(i).length(); j++){ 9 if(j >= m || i >= words.get(j).length() || words.get(i).charAt(j) != words.get(j).charAt(i)){ 10 return false; 11 } 12 } 13 } 14 return true; 15 } 16 }