leetcode 23: Longest Substring Without Repeating Characters

最新推荐文章于 2025-08-23 19:36:46 发布

转载最新推荐文章于 2025-08-23 19:36:46 发布 · 58 阅读

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原文链接：http://www.cnblogs.com/xishibean/archive/2013/01/09/2951400.html

文章标签：

#java #c/c++

本文介绍了一种算法，用于找出给定字符串中最长的不包含重复字符的子串长度。通过三种不同的实现方式（C++, Java 和 C++），详细展示了如何使用滑动窗口和哈希表等数据结构来高效解决此问题。

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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.

mistakes I made:1. forget record start point. I resetted longest to 0 after accountting a repeatted char which is wrong. It should start from the next char following the first char of two repeatted chars.

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if( s.size() == 0) return 0;
        
        #define MAX_LEN 256
        char flag[MAX_LEN];
        reset(flag, MAX_LEN);
        
        int max = 0;
        int longest = 0;
        int start = 0;
        
        for(int i=0; i<s.size(); i++) {
            if( flag[ s[i] ] == -1 ){
                longest++;
            } else {
                max = max < longest ? longest : max;
                
                for(int j=start; j<i; j++){
                    flag[ s[j] ] = -1;
                    if( s[j] == s[i] ) {
                        start = j + 1;
                        break;
                    }  
                }
                longest = i - start + 1;
                
            }
            
            flag[ s[i] ] = 1;
            
        }
        max = max < longest ? longest : max;
        return max;
    }
    
private:
    void reset(char a[], int n){
        for(int i=0; i<n; i++) {
            a[i] = -1;
        }
    } 

};

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int max = Integer.MIN_VALUE;
        int start = 0;
        HashMap<Character,Integer> map = new HashMap<Character, Integer>();
        
        for(int i=0; i<s.length(); i++) {
            char c = s.charAt(i);
            if( map.containsKey(c)){
                for(int j=start;j<i;j++) {
                    if(s.charAt(j)==c) break;
                    map.remove(s.charAt(j));
                }
                start = map.get(c)+1;
                map.put(c,i);
            } else {
                map.put(c,i);
                if( i-start+1 > max ) max = i-start+1;
            }
        }
        return max;
    }
}

c++:

class Solution {
public:


    int lengthOfLongestSubstring(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        unordered_set<char> myset;
        int sz = s.size();
        
        int cur = 0;
        int max = 0;
        int start = 0;
        
        for(int i=0; i<sz; i++) {
            char c = s[i];
            
            if( myset.find(c) == myset.end() ) {
                myset.insert(c);
                cur++;
                max = max>cur ? max : cur;
            } else {
                int j=start;
                while( s[j]!=c) {
                    myset.erase( s[j] );
                    ++j;
                }
                start = j+1;
                cur = i-start+1;
            }
        }
        return max;
    }
};

转载于:https://www.cnblogs.com/xishibean/archive/2013/01/09/2951400.html