标签:while turn new nat stringbu equal div empty therefore
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:
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.
public class Solution { public boolean validWordSquare(List<String> words) { if(words.isEmpty() || words.size() == 0) return true; int len = words.size(); for(int i = 0 ; i < len ; i++){ StringBuilder s1 = new StringBuilder(); String s2 = words.get(i); for(int j = 0 ; j < s2.length() ; j++){ if(j >= words.size()) return false; if(words.get(j).length() <= i) return false; s1.append(words.get(j).charAt(i)); } if(!s1.toString().equals(s2)) return false; } return true; } }
标签:while turn new nat stringbu equal div empty therefore
原文地址:http://www.cnblogs.com/joannacode/p/6096734.html