3. Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

my code:

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        set<char> seen;
        int ans = 0;
        for (int i = 0; i < s.length(); ++i) {
            int num = 0;
            for (int j  = i; j < s.length(); ++j) {
                if (seen.count(s[j]))
                    break;
                seen.insert(s[j]);
                num++;
            }
            seen.clear();
            ans = max(num, ans);
        }
        return ans;
    }
};

note: Runtime: 316 ms, faster than 5.95% of C++ online submissions for Longest Substring Without Repeating Characters.

 

effection code:

class Solution {
    public:
        int lengthOfLongestSubstring(string s) {
            if(s.size()<2) return s.size();
            int d=1, maxLen=1;
            unordered_map<char,int> map;
            map[s[0]]=0;
            for(int i=1;i<s.size();i++)
            {
                if(map.count(s[i])==0 || map[s[i]]<i-d)
                    d++;
                else
                    d= i- map[s[i]];
                map[s[i]]=i;
                if(d>maxLen)
                    maxLen = d;
            }
            return maxLen;
        }
    };

Runtime: 40 ms, faster than 24.00% of C++ online submissions for Longest Substring Without Repeating Characters.

 

第一种方法是最朴素的办法O(N^2)，第二种算法比第一种算法，效率高些O(N),再向前查找的时候用map保存了上次出现那个字母的位置，所以，只用线性的时间，就能够查找玩所有的元素。