LeetCode 854. K-Similar Strings

原题链接在这里：https://leetcode.com/problems/k-similar-strings/

题目：

Strings s1 and s2 are k-similar (for some non-negative integer k) if we can swap the positions of two letters in s1 exactly k times so that the resulting string equals s2.

Given two anagrams s1 and s2, return the smallest k for which s1 and s2 are k-similar.

Example 1:

Input: s1 = "ab", s2 = "ba"
Output: 1

Example 2:

Input: s1 = "abc", s2 = "bca"
Output: 2

Constraints:

• 1 <= s1.length <= 20
• s2.length == s1.length
• s1 and s2 contain only lowercase letters from the set {'a', 'b', 'c', 'd', 'e', 'f'}.
• s2 is an anagram of s1.

题解：

The smallest swap, we could use BFS.

For a current string, if cur.equals(s2), return returns current level.

If not, we get all its neighbors. 

To get a neighbor, we first find the different char between cur and s2, mark its index as ind. 

Then find the next char where cur.charAt(j) == s2.charAt(ind).

Time Complexity: O(k*n^2). n = s1.length(). k = swap count, could be up to n. There could be k level BFS iteration. In each iteration, for each cur, we could find n neighbor, totally there could be n node in the que, thus there are n^2 neighbors.

Space: O(n^2).

AC Java:

 1 class Solution {
 2     public int kSimilarity(String s1, String s2) {
 3         LinkedList<String> que = new LinkedList<>();
 4         HashSet<String> visited = new HashSet<>();
 5         que.add(s1);
 6         visited.add(s1);
 7         int level = 0;
 8         
 9         while(!que.isEmpty()){
10             int size = que.size();
11             while(size-- > 0){
12                 String cur = que.poll();
13                 if(cur.equals(s2)){
14                     return level;
15                 }
16                 
17                 for(String can : getNei(cur, s2)){
18                     if(visited.contains(can)){
19                         continue;
20                     }
21                     
22                     que.add(can);
23                     visited.add(can);
24                 }
25             }
26             
27             level++;
28         }
29         
30         return -1;
31     }
32     
33     private List<String> getNei(String s1, String s2){
34         char [] s1Arr = s1.toCharArray();
35         char [] s2Arr = s2.toCharArray();
36         int n = s1.length();
37         int ind = 0;
38         while(ind < n && s1Arr[ind] == s2Arr[ind]){
39             ind++;
40         }
41         
42         List<String> res = new ArrayList<>();
43         for(int j = ind + 1; j < n; j++){
44             if(s1Arr[j] == s2Arr[j] || s1Arr[j] != s2Arr[ind]){
45                 continue;
46             }
47             
48             swap(s1Arr, ind, j);
49             res.add(new String(s1Arr));
50             swap(s1Arr, ind, j);
51         }
52         
53         return res;
54     }
55     
56     private void swap(char [] arr, int i, int j){
57         char temp = arr[i];
58         arr[i] = arr[j];
59         arr[j] = temp;
60     }
61 }

标签：854,cur,int,s2,s1,ind,Similar,LeetCode,String
来源： https://www.cnblogs.com/Dylan-Java-NYC/p/16634647.html