[LeetCode]Word Squares

题目描述：

LeetCode 425. Word Squares

Given a set of words (without duplicates), find all word squares you can build from them.

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).

For example, the word sequence ["ball","area","lead","lady"] forms a word square because each word reads the same both horizontally and vertically.

b a l l
a r e a
l e a d
l a d y

Note:

1. There are at least 1 and at most 1000 words.
2. All words will have the exact same length.
3. Word length is at least 1 and at most 5.
4. Each word contains only lowercase English alphabet a-z.

Example 1:

Input:
["area","lead","wall","lady","ball"]

Output:
[
  [ "wall","area","lead","lady"
  ],
  [ "ball","area","lead","lady"
  ]
]

Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).

Example 2:

Input:
["abat","baba","atan","atal"]

Output:
[
  [ "baba","abat","baba","atan"
  ],
  [ "baba","abat","baba","atal"
  ]
]

Explanation:
The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).

题目大意：

给定一组没有重复的单词，从中寻找所有可能的“单词平方”。

一组单词如果其第k行和第k列组成的单词相同，其中0 ≤ k < max(numRows, numColumns)，则称其为一个单词平方。

例如，单词序列["ball","area","lead","lady"]组成一个单词平方，因为每一行每一列的单词读起来都是一样的。

b a l l
a r e a
l e a d
l a d y

注意：

1. 单词个数在[1, 1000]之间
2. 所有单词的长度均相同
3. 单词长度在[1, 5]之间
4. 每个单词只包含小写字母a-z

解题思路：

深度优先搜索（DFS）+ 剪枝（Pruning）

首先构建一个单词前缀prefix->单词word的字典mdict

深度优先搜索search(word, line)，其中word是当前单词，line是行数

利用变量matrix记录当前已经生成的单词

前缀prefix = matrix[0..line][line]，如果prefix对应单词不存在，则可以剪枝

否则枚举mdict[prefix]，并递归搜索

Python代码：

class Solution(object):
    def wordSquares(self, words):
        """
        :type words: List[str]
        :rtype: List[List[str]]
        """
        m = len(words)
        n = len(words[0]) if m else 0
        mdict = collections.defaultdict(set)
        for word in words:
            for i in range(n):
                mdict[word[:i]].add(word)
        matrix = []
        ans = []
        def search(word, line):
            matrix.append(word)
            if line == n:
                ans.append(matrix[:])
            else:
                prefix = ''.join(matrix[x][line] for x in range(line))
                for word in mdict[prefix]:
                    search(word, line + 1)
            matrix.pop()
        for word in words:
            search(word, 1)
        return ans