LeetCode【3】Longest Substring Without Repeating Characters

题目

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.

简单的说就是求最长无重复子串。

思路

“滑动窗口”
比方说 abcabccc 当你右边扫描到abca的时候你得把第一个a删掉得到bca，然后”窗口”继续向右滑动，每当加到一个新char的时候，左边检查有无重复的char，然后如果没有重复的就正常添加， 有重复的话就左边扔掉一部分（从最左到重复char这段扔掉），在这个过程中记录最大窗口长度

代码

public class LongestSubstring {public static int lengthOfLongestSubString(String s){    HashMap<Character, Integer> map = new HashMap<>();    int leftBound = 0;    int max = 0;    for(int i = 0;i<s.length();i++){        char c = s.charAt(i);        //窗口左边可能为下一个char，或者不变        leftBound = Math.max(leftBound, (map.containsKey(c))?map.get(c)+1:0);        max = Math.max(max, i-leftBound+1);//当前窗口长度        map.put(c, i);    }    return max;}}

完整运行代码

package LongestSubstring_3;import java.util.HashMap;public class LongestSubstring {public static int lengthOfLongestSubString(String s){    HashMap<Character, Integer> map = new HashMap<>();    int leftBound = 0;    int max = 0;    for(int i = 0;i<s.length();i++){        char c = s.charAt(i);        //窗口左边可能为下一个char，或者不变        leftBound = Math.max(leftBound, (map.containsKey(c))?map.get(c)+1:0);        max = Math.max(max, i-leftBound+1);//当前窗口长度        map.put(c, i);    }    return max;}public static void main(String[] args) {    String string = "bbabcdb";//abcabcbb,bbabcdb,abba    System.out.println(lengthOfLongestSubString(string));}}