lc425

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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 kth 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.

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

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

Example 1:

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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:

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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).

先构建一个字典树再用回溯法搜索。Trie我之前还没有写过，只知道概念，其结构是有一个26个nullptr的children指针的vector，有子树时那条边的字母对应的children元素从nullptr改为子树地址。另外还有index数组保存该节点子树中所有字符串在原words中对应的index。

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class TrieNode {
public:
    vector<int> index;
    vector<TrieNode*> children;
public:
    TrieNode():children(26, nullptr) { }
};
class Solution {
public:
    vector<vector<string>> res;
    TrieNode* buildTrie(vector<string>& words) {
        TrieNode* root = new TrieNode();
        TrieNode* node;
        for (int j = 0; j < words.size(); j++) {
            node = root;
            string word = words[j];
            for (int i = 0; i < word.length(); i++) {
                if (!node -> children[word[i]-'a']) {
                    TrieNode* new_node = new TrieNode();
                    node -> children[word[i]-'a'] = new_node;
                }
                node = node -> children[word[i]-'a'];
                node -> index.push_back(j);
            }
        }
        return root;
    }
    
    void backtracking(TrieNode* r, int level, vector<string> &square, vector<string> &words) {
        if (level >= words[0].length()) {
            res.push_back(square);
            return;
        }
        TrieNode* t = r;
        for (int i = 0; i < level; i++) {
            if (!t->children[square[i][level]-'a'])  return;
            t = t->children[square[i][level]-'a'];
        }
        for (auto in:t->index) {
            square[level] = words[in];
            backtracking(r, level+1, square, words);
            square[level] = "";
        }
        return;
    }
    vector<vector<string>> wordSquares(vector<string>& words) {
        vector<string> square{words.size(), ""};
        TrieNode* root = buildTrie(words);
        for (auto word: words) {
            square[0] = word;
            backtracking(root, 1, square, words);
        }
        return res;
    }
};