【LeetCode OJ 003】Longest Substring Without Repeating Characters
2015-11-05 20:27
417 查看
题目链接:https://leetcode.com/problems/longest-substring-without-repeating-characters/
题目: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.
解题思路:维护一个窗口,每次关注窗口中的字符串,维护一个HashSet,每扫描一个字符,如果HashSet中包含该字符,则移动左边窗口至重复字符的下一个字符,如果HashSet中没有改字符,则移动右边窗口。时间负复杂度为O(n),空间复杂度也为O(n)。具体代码如下:
public class Solution
{
public static void main(String[] args)
{
String str="abcbbc";
System.out.println(lengthOfLongestSubstring(str));
}
public static int lengthOfLongestSubstring(String s)
{
if(s==null && s.length()==0)
return 0;
HashSet<Character> set = new HashSet<Character>();
int max = 0; //最大不重复字符串长度
int left = 0; //不重复字符串起始位置
int right = 0; //不重复字符串结束为止
while(right<s.length())
{
//如果集合中包含当前所指的字符
if(set.contains(s.charAt(right)))
{
//获取当前不重复字符串的长度,如果大于max,则将max值替换
if(max<right-left)
{
max = right-left;
}
//将left移动到第一个重复的字符后面
while(s.charAt(left)!=s.charAt(right))
{
set.remove(s.charAt(left));
left++;
}
left++;
}
else
{
set.add(s.charAt(right));
}
right++;
}
max = Math.max(max,right-left);
return max;
}
}
题目: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.
解题思路:维护一个窗口,每次关注窗口中的字符串,维护一个HashSet,每扫描一个字符,如果HashSet中包含该字符,则移动左边窗口至重复字符的下一个字符,如果HashSet中没有改字符,则移动右边窗口。时间负复杂度为O(n),空间复杂度也为O(n)。具体代码如下:
public class Solution
{
public static void main(String[] args)
{
String str="abcbbc";
System.out.println(lengthOfLongestSubstring(str));
}
public static int lengthOfLongestSubstring(String s)
{
if(s==null && s.length()==0)
return 0;
HashSet<Character> set = new HashSet<Character>();
int max = 0; //最大不重复字符串长度
int left = 0; //不重复字符串起始位置
int right = 0; //不重复字符串结束为止
while(right<s.length())
{
//如果集合中包含当前所指的字符
if(set.contains(s.charAt(right)))
{
//获取当前不重复字符串的长度,如果大于max,则将max值替换
if(max<right-left)
{
max = right-left;
}
//将left移动到第一个重复的字符后面
while(s.charAt(left)!=s.charAt(right))
{
set.remove(s.charAt(left));
left++;
}
left++;
}
else
{
set.add(s.charAt(right));
}
right++;
}
max = Math.max(max,right-left);
return max;
}
}
相关文章推荐
- poj1163 dp the triangle
- HDU 5531 Rebuild (2015长春现场赛,计算几何+三分法)
- jsp页面弹出For input string:""
- 嵌入式 多线程下变量-原子操作__sync_fetch_and_add等等
- spring的四种注入方式
- 2015 11 04 函数指针
- IOS开发 - KVC和KVO的使用
- Codeforces 593 A. 2Char 【Codeforces Round #329 (Div. 2)】
- Mysql ERROR 1045 (28000): Access denied for user 'root'@'localhost'问题的解决
- 多系统PE win UBUNTU OSX
- 2015-10-28 C#4
- lintcode 中等题:Palindrome Linked List 回文链表
- poj 1011
- cjson不连续存储问题
- 字符串回文子序列问题
- 2015 11 03 内存分区
- [JPA错误]javax.persistence.EntityNotFoundException: Unable to find xxx
- 第9章 保护spring应用 9.1 Spring Security介绍
- Application, Activity, ContentProvider启动顺序
- synchronized详解