LeetCode-Longest Substring Without Repeating Characters<ERROR>

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disappearedgod

最新推荐文章于 2025-05-18 01:04:12 发布

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分类专栏：字符串(String) LeetCode 文章标签： leetcode String

本文链接：https://blog.youkuaiyun.com/disappearedgod/article/details/38824871

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本文探讨如何在给定字符串中找到最长的无重复字符子串，通过Trie树法和后缀数组方法实现解决方案。

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作者：disappearedgod

文章出处：http://blog.youkuaiyun.com/disappearedgod/article/details/38824871

时间：2014-8-25

题目

Longest Substring Without Repeating Characters

Total Accepted: 20666 Total Submissions: 92944 My Submissions

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

想法

举一个例子“abcabcabcd”

第一种方法 Trie树法

将String的后缀序列输入Trie树。

public class Solution {
    private TrieST trie;
    public int lengthOfLongestSubstring(String s) {
        int res = 0;
        int tmp = 0;
        HashSet<Integer> set = new HashSet<Integer>();
        if(s.length() == 0)
            return 0;
        trie = new TrieST();
        for(int i = 0; i < s.length(); i++){
            trie.put(s.substring(i));
            tmp = set.get(trie.get(s.substring(i)));
            res = res > tmp ? res : tmp;
        }
        return res;  
    }
    public class TrieST{
        private Node root;
        private int R = 256;
        private class Node{
            private int val;
            private Node[] next = new Node[R];
        }
        public int get(String key){
            Node x = get(root, key, 0);
            if(x == null)
                return null;
            return x.val;
        }
        private int get(Node x, String key, int d){
            if(x == null) return null;
            if(d == key.length()) return x;
            char c = key.charAt(d);
            return get(x.next[c], key, d+1);
        }
        
        public void put(String key, int val){
            root = put(root, key, val, 0);
        }
        
        private Node put(Node x, String key, int val, int d){
            if(x == null) return null;
            if(d == key.length()){
                x.val = val;
                return x;
            }
            char c = key.charAt(d);
            x.next[c] = put(x.next[c], key, val, d+1);
        }
    }
   
}

暴力法后缀数组

想用一下JAVAAPI解决一下后缀数组题，用list和Collection排序。

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        ArrayList<String> a = new ArrayList<String>();
        for(int i = 0; i < s.length(); i++)
            a.add(s.substring(i));
        Collections.sort(a);
        int max = 0;
        String ret = "";
        for(int i = 1; i < s.length(); i++){
            if(max < a.get(i).compareTo(a.get(i-1))){
                max =a.get(i).compareTo(a.get(i-1));
                ret = a.get(i-1);
            }
        }
        return max;
    }
}

结果