p91 无重复字符的最长子串(leetcode 3)

一：解题思路

方法一：Time:O(n),Space:O(1),方法一比方法二容易理解,推荐用方法一

方法二：Time:O(n),Space:O(1)

二：完整代码示例 (C++版和Java版)

方法一C++:

class Solution {
public:
    int lengthOfLongestSubstring(string s) 
    {
        if (s.size() == 0) return 0;
        vector<int> count(256,0);
        int maxLen = 0;
        int i = 0;
        int j = 0;
        for (; i < s.size(); i++)
        {
            for (; j < s.size(); j++)
            {
                if (count[s[j]] != 0) break;
                count[s[j]] += 1;
            }

            maxLen = max(maxLen,j-i);
            count[s[i]] -= 1;
        }

        return maxLen;
    }
};

方法二C++:

class Solution {
public:
    int lengthOfLongestSubstring(string s) 
    {
        if (s.size() == 0) return 0;
        int maxLen = 0;
        vector<int> index(256,-1);
        for (int i=0,j = 0; j < s.size(); j++)
        {
            i = max(index[s[j]]+1,i);
            maxLen = max(maxLen,j-i+1);
            index[s[j]] = j;
        }

        return maxLen;
    }
};

方法一Java:

class Solution {
        public int lengthOfLongestSubstring(String s) 
        {
               if(s==null || s.length()==0) return 0;
               int[] count=new int[256];
               int maxLen=0;
               int i=0;
               int j=0;
               
               for(;i<s.length();i++)
               {
                   for(;j<s.length();j++)
                   {
                       if(count[s.charAt(j)]!=0) break;
                       count[s.charAt(j)]+=1;
                   }
                   
                   maxLen=Math.max(maxLen,j-i);
                   count[s.charAt(i)]-=1;
               }
               
               return maxLen;
        }
    }

方法二Java:

class Solution {
        public int lengthOfLongestSubstring(String s) 
        {
               if(s==null || s.length()==0) return 0;
               int[] index=new int[256];
               Arrays.fill(index,-1);
               int maxLen=0;
               for(int i=0,j=0;j<s.length();j++)
               {
                   i=Math.max(index[s.charAt(j)]+1,i);
                   maxLen=Math.max(maxLen,j-i+1);
                   index[s.charAt(j)]=j;
               }
               
               return maxLen;
        }
    }