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.
题意：给出一个字符串，找到最长不含重复字符的子串。
思路：设定两个下标i，j。i指向子串的第一个字符，j指向子串最后一个字符。i和j初始时都指向父串的第一个字符，j逐个向后移动，并且同时维护maxLength，当遇到某个字符与之前的第k个字符相同，则将i移动到第k+1个。虽然有两个while循环，但是算法复杂度仍为O(n)。
public int lengthOfLongestSubstring(String s) { if(s.equals("")){ return 0; } int maxLength = 1;//排除了空串的情况下，最小长度就为1 int i = 0; int j = 0; boolean[] map = new boolean[300];//标记数组 char[] charArray = s.toCharArray();//转换为字符数组 map[charArray[0]] = true;//第一个先设为有 while(j < s.length() - 1){ j++; while (map[charArray[j]]){//当j遇到某个之前已经出现过的字符，则i++只到跳过那个字符 map[charArray[i]] = false; i++; } map[charArray[j]] = true;//子序列中新增一个字符 maxLength = Math.max(maxLength, j - i + 1);//维护maxLength } return maxLength; }
