Problem:
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.
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:
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).
Solution:
又是道恐怖的Trie的题目，这道题首先需要创建Trie数据结构，里面有个函数getStartsWith(TrieNode *r)是用DFS的方法获取所有从节点r开始的后缀。下面用Example1说明算法思路，path初始化为长宽均为4的字符串数组，首先遍历所有字符串先放满第一行和第一列，比如：
["wall",
"a ",
"l ",
"l "]
然后开始放第二行和第二列，此时我们需要用getStartsWith(TrieNode *r)函数获取前缀为a的所有后缀组合，并一个个尝试，其中一个结果为：
["wall",
"area",
"le ",
"la "]
然后对于第三行和第三列找到前缀为le的所有后缀组合，以此类推，直到index为5即可。
Code:
1 struct TrieNode{
2 TrieNode *next[26];
3 bool isEnd;
4 TrieNode():isEnd(false){
5 for(int i = 0;i != 26;++i)
6 next[i] = NULL;
7 }
8 };
9 class Trie{
10 public:
11 Trie(){
12 root = new TrieNode();
13 }
14 void insert(string &word){
15 TrieNode *current = root;
16 for(int i = 0;i != word.size();++i){
17 if(current->next[word[i]-‘a‘] == NULL){
18 current->next[word[i]-‘a‘] = new TrieNode();
19 }
20 current = current->next[word[i]-‘a‘];
21 }
22 current->isEnd = true;
23 }
24 void dfs(TrieNode *r,string path,vector<string> &result){
25 if(r->isEnd){
26 result.push_back(path);
27 return;
28 }
29 for(int i = 0;i != 26;++i){
30 if(r->next[i] != NULL){
31 dfs(r->next[i],path+(char)(i+‘a‘),result);
32 }
33 }
34 }
35 vector<string> getStartsWith(TrieNode *r) {
36 vector<string> result;
37 dfs(r,"",result);
38 return result;
39 }
40 TrieNode *getRoot(){
41 return root;
42 }
43 private:
44 TrieNode *root;
45 };
46 class Solution {
47 public:
48 void backtrace(int index,Trie &tree,TrieNode *root,vector<string> &path,vector<vector<string>> &result){
49 int m = path.size();
50 if(index == m){
51 result.push_back(path);
52 return;
53 }
54 TrieNode *current = root;
55 for(int i = 0;i != index;++i){
56 if(current->next[path[index][i]-‘a‘] == NULL)
57 return;
58 current = current->next[path[index][i]-‘a‘];
59 }
60 vector<string> suffix = tree.getStartsWith(current);
61 for(int i = 0;i != suffix.size();++i){
62 for(int j = index;j != m;++j){
63 path[index][j] = suffix[i][j-index];
64 path[j][index] = suffix[i][j-index];
65 }
66 backtrace(index+1,tree,root,path,result);
67 for(int j = index;j != m;++j){
68 path[index][j] = ‘ ‘;
69 path[j][index] = ‘ ‘;
70 }
71 }
72 }
73 vector<vector<string>> wordSquares(vector<string>& words) {
74 Trie tree;
75 for(int i = 0;i != words.size();++i)
76 tree.insert(words[i]);
77 TrieNode *root = tree.getRoot();
78 int m = words[0].size();
79 vector<vector<string>> result;
80 vector<string> path(m,string(m,‘ ‘));
81 for(int i = 0;i != words.size();++i){
82 for(int j = 0;j != m;++j){
83 path[0][j] = words[i][j];
84 path[j][0] = words[i][j];
85 }
86 backtrace(1,tree,root,path,result);
87 for(int j = 0;j != m;++j){
88 path[0][j] = ‘ ‘;
89 path[j][0] = ‘ ‘;
90 }
91 }
92 return result;
93 }
94 };
原文地址：https://www.cnblogs.com/haoweizh/p/10204610.html