[LeetCode]422. Valid Word Square

最新推荐文章于 2023-02-25 22:41:55 发布

pablojie

最新推荐文章于 2023-02-25 22:41:55 发布

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文章标签： LeetCode java

本文链接：https://blog.youkuaiyun.com/pablojie/article/details/52868798

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本文介绍了一种算法，用于检查一组单词是否构成有效的单词方阵。通过对比每一对单词的相应字符来验证，提供了两种实现方法并讨论了边界条件。

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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.
题目大意：(吐槽一下，这题目真是又臭又长)这道题的意思是检测第i行中第j个字符是否等于第j行的第i个字符，从题目的意思其实我们用两个for循环遍历就能解决。
个人觉得这道题真正的考点是给你的第二个例子，就是各种边界条件你是否能考虑全面。我个人看下来需要考虑的有这么几个点：
1.如果某个string的长和list的size不同，那么返回false
2.如果搜索的某一个字符所在的index发生StringIndexOutOfBoundsException或者IndexOutOfBoundsException时就返回false
3.最后当然是最重要的就是如果第i行第j个字符与第j行第i个字符不相等，那么就返回false
public class Solution {
    public boolean validWordSquare(List<String> words) {
        try{
            for(int row = 0; row <words.size();row++){
                String word = words.get(row);
                for(int col = 0;col<word.length();col++){
                    String word_cmp = words.get(col);
                    if(word.charAt(col) != word_cmp.charAt(row)){
                        return false;
                    }
                }
            }
        }catch(StringIndexOutOfBoundsException e){
            return false;
        }catch(IndexOutOfBoundsException ee){
            return false;
        }
        return true;
    }
}
然后在讨论区看到了一个非常好的思路，就是说先把每一行转化为一个string，然后用行和列来进行对比，这样就不需要考虑那么多的边界条件。
这样一个equals方法就能让我们得到结果。
public class Solution {
    public boolean validWordSquare(List<String> words) {
        
         for(int i=0;i<words.size();i++){
             String s = words.get(i);
             if(!s.equals(getVertical(i,words))){
                 return false;
             }
         }
         return true;
    }
    
    private String getVertical(int col,List<String> words){
        StringBuilder sb = new StringBuilder();
        for(int i=0;i<words.size();i++){
            String word = words.get(i);
            if(col<word.length()){
                sb.append(word.charAt(col));
            }
        }
        return sb.toString();
    }
    
    
}