Longest Substring Without Repeating Characters

Longest Substring Without Repeating Characters

问题：

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.

思路：

　　two point + hashtable

我的代码：（未通过，超时，(⊙﹏⊙)b）

import java.util.Hashtable;
public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0)    return 0;
        Hashtable<Character,Integer> ht = new Hashtable<Character,Integer>();
        int rst = 0;
        int max = 0;
        int i = 0;
        while(i < s.length())
        {
            char c = s.charAt(i);
            if(!ht.containsKey(c))
            {
                ht.put(c, i);
                max++;
                rst = Math.max(max,rst);
                i++;
            }
            else
            {
                int pastKey = ht.get(c);
                ht.clear();
                max = 0;
                i = pastKey + 1;
            }
        }
        return rst;
    }
}

 他人代码1：

import java.util.Hashtable;
public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(s == null || s.length() == 0)    return 0;
        Hashtable<Character,Integer> ht = new Hashtable<Character,Integer>();
        int max = 0;
        int left = 0;
        for(int right = 0; right < s.length(); right++)
        {
            char c = s.charAt(right);
            if(ht.containsKey(c) && ht.get(c) >= left)
            {
                left = ht.get(c) + 1;
            }
            ht.put(c, right);
            max = Math.max(max, right - left + 1);
        }
        return max;
    }
}

他人代码2：（用数组代替HashMap）

public int lengthOfLongestSubstring(String s) {
        if (s == null) {
            return 0;
        }
        
        int max = 0;
        
        // suppose there are only ASCII code.
        int[] lastIndex = new int[128];
        for (int i = 0; i < 128; i++) {
            lastIndex[i] = -1;
        }
        
        int len = s.length();
        int l = 0;
        for (int r = 0; r < len; r++) {
            char c = s.charAt(r);
            
            if (lastIndex[c] >= l) {
                l = lastIndex[c] + 1;
            }
            
            // replace the last index of the character c.
            lastIndex[c] = r;
            
            // replace the max value.
            max = Math.max(max, r - l + 1);
        }
        
        return max;
    }

学习之处：

• 想的太复杂，没有抓住问题的实质，平常写代码也是，先考虑!ht.containsKey() 是不是也可以先考虑 ht.containsKey的情况呢。
• ASCLL是一个字节，是8位，而最高位是用来进行奇偶校验的，所以可以表示的数字有2⁷=128 故可以使用128的数组进行存储，代替HashMap。
• 双指针体现在，一个在前走着，确定上界（上限），后面的那个用来确定下界（下限） 此为上下界中的双指针问题。