Given a string, find the length of the longest substring without repeating characters.
Example:
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.
我自己写的一个方法是爆搜,但是这个方法超时了。
代码如下:
package com.leetcode.study; import java.util.HashMap; import java.util.HashSet; import java.util.Map; import java.util.Set; public class Main { public static void main(String[] args) { System.out.println(lengthOfLongestSubstring("wllxdiklosdrdxfohgwringzefwbytmwgxtjhdxwycpbawphcnbmajmeokhoftlmsexakuyixplxmagoojdospvjbcxh")); } public static int lengthOfLongestSubstring(String s){ Map<String,Integer> map = new HashMap<String, Integer>(); for(int i = 0;i<s.length();i++){ if(map.containsKey(s.charAt(i)+"")){ continue; }else{ map.put(s.charAt(i)+"", i); } } Set<String> set; int length = 0; boolean flag = false; for(int i = map.size();i>0;i--){ for (int j = 0; j < s.length()-i+1; j++) { String temp = s.substring(j, j+i); set = new HashSet<String>(); for(int k=0;k<temp.length();k++){ set.add(temp.charAt(k)+""); } if(set.size()==temp.length()){ length=set.size(); flag=true; break; } } if(flag){ break; } } return length; } }
代码运行超时。
还是看看别人的代码吧。
滑动窗口算法:
package com.leetcode.study; import java.util.HashMap; import java.util.HashSet; import java.util.Map; import java.util.Set; public class Main { public static void main(String[] args) { System.out.println(lengthOfLongestSubstring("abcabcbb")); } public static int lengthOfLongestSubstring(String s){ int n = s.length(); Set<Character> set = new HashSet<Character>(); int ans=0,i=0,j=0; while(i<n&&j<n){ if(!set.contains(s.charAt(j))){ set.add(s.charAt(j)); j++; ans=Math.max(ans, j-i); }else{ set.remove(s.charAt(i)); i++; } } return ans; } }
这个方法没太弄明白。