给定一个单词集合 words (没有重复),找出并返回其中所有的 单词方块 。 words 中的同一个单词可以被 多次 使用。你可以按 任意顺序 返回答案。
一个单词序列形成了一个有效的 单词方块 的意思是指从第 k 行和第 k 列 0 <= k < max(numRows, numColumns) 来看都是相同的字符串。
例如,单词序列 ["ball","area","lead","lady"] 形成了一个单词方块,因为每个单词从水平方向看和从竖直方向看都是相同的。
输入:words = ["area","lead","wall","lady","ball"]
输出: [["ball","area","lead","lady"],
["wall","area","lead","lady"]]
解释:
输出包含两个单词方块,输出的顺序不重要,只需要保证每个单词方块内的单词顺序正确即可。
输入:words = ["abat","baba","atan","atal"]
输出:[["baba","abat","baba","atal"],
["baba","abat","baba","atan"]]
解释:
输出包含两个单词方块,输出的顺序不重要,只需要保证每个单词方块内的单词顺序正确即可。
- 1 <= words.length <= 1000
- 1 <= words[i].length <= 4
- words[i] 长度相同
- words[i] 只由小写英文字母组成
- words[i] 都 各不相同
use std::collections::HashMap;
impl Solution {
fn word_squares(words: Vec<String>) -> Vec<Vec<String>> {
let n = words.len();
let m = words[0].len();
let words: Vec<Vec<char>> = words.into_iter().map(|s| s.chars().collect()).collect();
let mut prefix: HashMap<Vec<char>, Vec<&[char]>> = HashMap::new();
for word in words.iter() {
for i in 0..=m {
prefix.entry(word[..i].to_vec()).or_default().push(word);
}
}
let mut ans = vec![];
Self::dfs(&prefix, &mut ans, &mut Vec::new(), n, m, 0);
ans
}
fn dfs<'a>(prefix: &HashMap<Vec<char>, Vec<&'a [char]>>, ans: &mut Vec<Vec<String>>, selected: &mut Vec<&'a [char]>, n: usize, m: usize, begin: usize) {
if begin == m {
ans.push((0..m).map(|i| selected[i].iter().collect()).collect());
} else {
let mut tmp = vec![];
for i in 0..begin {
tmp.push(selected[i][begin]);
}
if let Some(candidates) = prefix.get(&tmp) {
for word in candidates {
selected.push(word);
Self::dfs(prefix, ans, selected, n, m, begin + 1);
selected.pop();
}
}
}
}
}