leetcode-3. Longest Substring Without Repeating Characters

前言：最近开学事情特别多，隔了一个月没刷题，真是蛋疼。

----------------------

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.

首先想到的最简单的O(n^2)的时间效率代码，提交之后果然超时。

public class Solution {
public int lengthOfLongestSubstring(String s) {
int maxLength = 0;
for(int i = 0 ; i < s.length() - maxLength; i ++){

HashMap<Character,Integer> s_map = new HashMap<Character,Integer>();
for(int j = i; j < s.length(); j ++){
if(s_map.containsKey(s.charAt(j))) break;
else s_map.put(s.charAt(j),j);
}

if(s_map.size() > maxLength) maxLength = s_map.size();
}
return maxLength;
}
}

稍微进行了改动，当碰到第一个重复的字符串的时候，将i的值扩大。但是效率还是比较低，You are here!

Your runtime beats 3.07% of java submissions.。效果有点差，后续再看怎么改进

public class Solution {
public int lengthOfLongestSubstring(String s) {
int maxLength = 0;
for(int i = 0 ; i < s.length() - maxLength; i ++){

HashMap<Character,Integer> s_map = new HashMap<Character,Integer>();
for(int j = i; j < s.length(); j ++){
if(s_map.containsKey(s.charAt(j))) {i = s_map.get(s.charAt(j));break;}
else s_map.put(s.charAt(j),j);
}

if(s_map.size() > maxLength) maxLength = s_map.size();
}
return maxLength;
}
}