3.Longest Substring Without Repeating Characters

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本文介绍了一种高效算法，用于解决字符串中找到最长无重复字符子串的问题，通过设置数组来跟踪字符出现情况，实现O(n)的时间复杂度。

Longest Substring Without Repeating Characters

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.

solution: The traditional way to slove it is O(n^3). Our solution is O(n). Set an array a[300] to represent all character and set it to 0 for start. set i = 0. For each entry j in array, if the a[s.charAt(j)-' ']==0, it means s.charAt(j) has not appeared, and make a[s.charAt(j)-' ']=1 , store the current len j-i+1. and move on. if the a[s.charAt(j)-' ']==1 which tell us it repeatd. to reset the a from the position i, set the a[s.charAt(i)-' ']=1 until we remove the s.charAt(j). Then make the a[s.charAt(j)-' ']=1 to contiune find.

Note that we use array to for the current substring (between i and j), another way is use hashset.

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int []a = new int[300];
        for (int k=0;k<300;k++) {
            a[k] = 0;
        }
        int i,j;
        int len = 0;
        for (i=0,j=0;j<s.length();j++) {
            if (a[s.charAt(j)-' '] == 0) {
                a[s.charAt(j)-' '] = 1;
                if(j-i+1 >len) {
                    len = j-i+1;
                }
            }
            else {
                while (a[s.charAt(j)-' '] == 1) {
                    a[s.charAt(i)-' '] = 0;
                    i++;
                }
                a[s.charAt(j)-' '] = 1;
            }
        }
        return len;
    }
}