422. 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:

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",
  "read",
  "lady"
]

Output:
false

Explanation:
The third row reads "read" while the third column reads "lead".

Therefore, it is NOT a valid word square.

链接：https://leetcode.com/problems/valid-word-square/#/description

3/25/2017

注意：

只看给的几个例子是不行的，要把题目读全，

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).

说实话这道题是蒙对的，performance居然还不错26ms

public class Solution {
    public boolean validWordSquare(List<String> words) {
        int rowLength = words.size();
        for (String s: words) {
            if (s.length() > rowLength) return false;
        }
        for (int i = 0; i < rowLength; i++) {
            for (int j = i; j < rowLength; j++) {
                if (words.get(i).length() > j && words.get(j).length() > i && words.get(i).charAt(j) == words.get(j).charAt(i)) ;
                else if (words.get(i).length() <= j && words.get(j).length() <= i) ;
                else return false;
            }
        }
        return true;
    }
}

别人类似的算法

不同的是j的范围是根据当前i行的长度来的，判断条件里前两个是为了保证charAt()函数安全。

public class Solution {
    public boolean validWordSquare(List<String> words) {
        if(words == null || words.size() == 0){
            return true;
        }
        int n = words.size();
        for(int i=0; i<n; i++){
            for(int j=0; j<words.get(i).length(); j++){
                if(j >= n || words.get(j).length() <= i || words.get(j).charAt(i) != words.get(i).charAt(j))
                    return false;
            }
        }
        return true;
    }
}

一行Python

The map(None, ...) transposes the "matrix", filling missing spots with None.

def validWordSquare(self, words):
    t = map(None, *words)
    return t == map(None, *t)

更多讨论：

https://discuss.leetcode.com/category/551/valid-word-square