LeetCode 1044. Longest Duplicate Substring

2022-08-08 06:30:44 阅读：218 来源： 互联网

标签：substring hash int length len Substring start Duplicate Longest

原题链接在这里：https://leetcode.com/problems/longest-duplicate-substring/

题目：

Given a string s, consider all duplicated substrings: (contiguous) substrings of s that occur 2 or more times. The occurrences may overlap.

Return any duplicated substring that has the longest possible length. If s does not have a duplicated substring, the answer is "".

Example 1:

Input: s = "banana"
Output: "ana"

Example 2:

Input: s = "abcd"
Output: ""

Constraints:

2 <= s.length <= 3 * 104
s consists of lowercase English letters.

题解：

We could guess a length, and check if there is duplicate.

If there is, then we could guess a bigger one. Otherwise, a smaller one.

How to check duplicates with given length. Use rolling hash.

"abcd", len = 3. when "abc" we have a hash, when it comes to "d", first decrease "a" then add "d" hash.

Time Complexity: O(nlogn). n = s.length().

Space: O(n).

AC Java:

 1 class Solution {
 2     int m = 1000000007;
 3     int base = 26;
 4     
 5     public String longestDupSubstring(String s) {
 6         if(s == null || s.length() == 0){
 7             return s;
 8         }    
 9         
10         int [] nums = new int[s.length()];
11         for(int i = 0; i < s.length(); i++){
12             nums[i] = (int)(s.charAt(i) - 'a');
13         }
14         int l = 1;
15         int r = s.length();
16         while(l <= r){
17             int mid = l + (r - l) / 2;
18             if(isValid(s, mid, nums) != -1){ 
19                 l = mid + 1;
20             }else{
21                 r = mid - 1;
22             }
23         }
24         
25         int start = isValid(s, l - 1, nums);
26         return s.substring(start, start + l - 1);
27     }
28     
29     private int isValid(String s, int len, int [] nums){
30         int n = s.length();
31         Map<Long, List<Integer>> map = new HashMap<>();
32         long hash = 0;
33         for(int i = 0; i < len; i++){
34             hash = (hash * base + (s.charAt(i) - 'a')) % m;
35         }
36         
37         map.put(hash, Arrays.asList(0));
38         
39         // pre-compute deduction multiplier
40         long mul = 1;
41         for(int i = 0; i < len; i++){
42             mul = (mul * base) % m;
43         }
44         
45         for(int start = 1; start < n - len + 1; start++){
46             hash = (hash * base - nums[start - 1] * mul % m + m) % m;
47             hash = (hash + nums[start + len - 1]) % m;
48             List<Integer> list = map.get(hash);
49             if(list != null){
50                 // There are equal hash before
51                 String cur = s.substring(start, start + len);
52                 for(int ind : list){
53                     if(s.substring(ind, ind + len).equals(cur)){
54                         return ind;
55                     }
56                 }
57             }
58             
59             map.putIfAbsent(hash, new ArrayList<Integer>());
60             map.get(hash).add(start);
61         }
62         
63         return -1;
64     }
65 }

类似Longest Repeating Substring.

标签：substring,hash,int,length,len,Substring,start,Duplicate,Longest
来源： https://www.cnblogs.com/Dylan-Java-NYC/p/16560459.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。

ICode9

LeetCode 1044. Longest Duplicate Substring