Description

Given a set of strings which just has lower case letters and a target string, output all the strings for each the edit distance with the target no greater than k.

You have the following 3 operations permitted on a word:

  • Insert a character
  • Delete a character
  • Replace a character

Example

Example 1:

Given words = `["abc", "abd", "abcd", "adc"]` and target = `"ac"`, k = `1`
Return `["abc", "adc"]`
Input:
["abc", "abd", "abcd", "adc"]
"ac"
1
Output:
["abc","adc"] Explanation:
"abc" remove "b"
"adc" remove "d"

Example 2:

Input:
["acc","abcd","ade","abbcd"]
"abc"
2
Output:
["acc","abcd","ade","abbcd"] Explanation:
"acc" turns "c" into "b"
"abcd" remove "d"
"ade" turns "d" into "b" turns "e" into "c"
"abbcd" gets rid of "b" and "d"

思路:滚动数组。用字典树对dfs进行优化。
class TrieNode{
public TrieNode[] sons;
public boolean isWord;
public String word; public TrieNode() {
int i;
sons = new TrieNode[26];
for (i = 0; i < 26; ++i) {
sons[i] = null;
} isWord = false;
} static public void Insert(TrieNode p, String word) {
int i;
char[] s = word.toCharArray();
for (i = 0; i < s.length; ++i) {
int c = s[i] - 'a';
if (p.sons[c] == null) {
p.sons[c] = new TrieNode();
} p = p.sons[c];
} p.isWord = true;
p.word = word;
}
} public class Solution {
/**
* @param words: a set of stirngs
* @param target: a target string
* @param k: An integer
* @return: output all the strings that meet the requirements
*/ int K;
int n;
char[] target;
List<String> res; // p is the current TrieNode
// f[] representss f[Sp][...]
void dfs(TrieNode p, int[] f) {
int[] newf;
int i;
if (p.isWord && f[n] <= K) {
res.add(p.word);
} for (int c = 0; c < 26; ++c) {
if (p.sons[c] == null) {
continue;
} // calc newf
newf = new int[n + 1];
// newf[...]: f[Sp + c][....] // newf[j] = Math.min(Math.min(f[j], newf[j-1]), f[j-1]) + 1;
for (i = 0; i <= n; ++i) {
newf[i] = f[i] + 1;
} for (i = 1; i <= n; ++i) {
newf[i] = Math.min(newf[i], f[i - 1] + 1);
} for (i = 1; i <= n; ++i) {
if (target[i - 1] - 'a' == c) {
newf[i] = Math.min(newf[i], f[i - 1]);
} newf[i] = Math.min(newf[i - 1] + 1, newf[i]);
} dfs(p.sons[c], newf);
}
} public List<String> kDistance(String[] words, String targets, int k) {
res = new ArrayList<String>();
K = k;
TrieNode root = new TrieNode();
int i;
for (i = 0; i < words.length; ++i) {
TrieNode.Insert(root, words[i]);
} target = targets.toCharArray();
n = target.length;
int[] f = new int[n + 1];
for (i = 0; i <= n; ++i) {
f[i] = i;
} dfs(root, f);
return res;
}
}

  

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