3. Longest Substring Without Repeating Characters

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本文探讨了如何在给定字符串中寻找最长的无重复字符子串，通过示例解析了算法实现，如abcabcbb最长无重复子串为abc，长度为3。介绍了使用HashMap或Array来跟踪字符位置，实现O(n)复杂度的解决方案。

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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.

O(n) solution(思路:用hashmap可快速锁定当前字母是否有重复，由此可以确定当前[i,j]的i的值，由于已知仅有256个字符，用Array来做index会更快)

class Solution {
    public int lengthOfLongestSubstring(String s) {
        Map<Character,Integer> map = new HashMap<Character,Integer>();
        int max=0,n=s.length();
        for(int i =0,j=0 ;j<n; j++){
            if(map.containsKey(s.charAt(j))){
                i = Math.max(map.get(s.charAt(j))+1,i);
            }
            max = Math.max(max,j-i+1);
            map.put(s.charAt(j),j);
        }
        return max;
    }
}