3. Longest Substring Without Repeating Characters

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

Solution

(1) Java

public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        int rst = 1;
        int start = 0;
        char[] arr = s.toCharArray();
        Map<Character, Integer> map = new HashMap<>();
        map.put(arr[0],0);
        for (int i = 1; i < s.length(); i++) {
            char c = arr[i];
            if (map.containsKey(c) && map.get(c) >= start) {
                start = map.get(c)+1;
            } 
            map.put(c, i);
            rst = Math.max(rst, i-start+1);
        }       
        return rst;
    }

(2) Python

def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        if not s:
            return 0
        rst = 1
        cache = {s[0]:0}
        start = 0
        for i in range(1, len(s)):
            c = s[i]
            if c in cache and cache[c] >= start:
                start = cache[c]+1
            cache[c]=i
            rst = max(rst, i-start+1)
        return rst

(3) Scala

object Solution {
    def lengthOfLongestSubstring(s: String): Int = {
        var start:Int = 0
        var idxMap:Map[Char, Int] = Map()
        var rst:Int = 0
        for (i <- 0 until s.length()) {
            var c = s.charAt(i)
            if (idxMap.contains(c) && idxMap(c) >= start) {
                rst = rst.max(i-start)
                start = idxMap(c)+1
            }
            idxMap += (c -> i)
        }
        rst = rst.max(s.length()-start)
        return rst
    }
}