*387. First Unique Character in a String (linkedhashmap + string) debug manunally instead of using leetcode
The ability to debug and overall thinking need to improve
Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
Examples:
s = "leetcode"
return 0. s = "loveleetcode",
return 2.
Note: You may assume the string contain only lowercase letters.
Solution: using linkedHashMap<> : have the insertion order!
class Solution {
public int firstUniqChar(String s) {
Map<Character, Integer> map = new LinkedHashMap<Character, Integer>();
for(int i = 0; i<s.length(); i++){
Character c = s.charAt(i);
if(map.containsKey(c)){
map.put(c, 2);
}else {
map.put(c,1);
}
}
char temp = '#';
for(Map.Entry<Character, Integer> entry : map.entrySet()){
System.out.println(entry.getKey() + " " + entry.getValue());
if(entry.getValue() == 1){
temp = entry.getKey();
break;
}
}
for(int i = 0; i<s.length(); i++){
if(s.charAt(i) == temp) return i;
}
return -1;
}
}
*387. First Unique Character in a String (linkedhashmap + string) debug manunally instead of using leetcode的更多相关文章
- 【LeetCode】387. First Unique Character in a String
Difficulty:easy More:[目录]LeetCode Java实现 Description https://leetcode.com/problems/first-unique-cha ...
- LeetCode 387. First Unique Character in a String (字符串中的第一个唯一字符)
题目标签:String, HashMap 题目给了我们一个 string,让我们找出 第一个 唯一的 char. 设立一个 hashmap,把 char 当作 key,char 的index 当作va ...
- 387. First Unique Character in a String
Given a string, find the first non-repeating character in it and return it's index. If it doesn't ex ...
- LeetCode 387. First Unique Character in a String
Problem: Given a string, find the first non-repeating character in it and return it's index. If it d ...
- leetcode修炼之路——387. First Unique Character in a String
最近公司搬家了,有两天没写了,今天闲下来了,继续开始算法之路. leetcode的题目如下: Given a string, find the first non-repeating characte ...
- 18. leetcode 387. First Unique Character in a String
Given a string, find the first non-repeating character in it and return it's index. If it doesn't ex ...
- [LeetCode&Python] Problem 387. First Unique Character in a String
Given a string, find the first non-repeating character in it and return it's index. If it doesn't ex ...
- [leetcode]387. First Unique Character in a String第一个不重复字母
Given a string, find the first non-repeating character in it and return it's index. If it doesn't ex ...
- Java [Leetcode 387]First Unique Character in a String
题目描述: Given a string, find the first non-repeating character in it and return it's index. If it does ...
随机推荐
- Docker 使用samba 共享文件
Docker 使用samba 共享文件 docker run -it --name samba \ -p 139:139 -p 445:445 \ -v /home/develop/code/de ...
- 经典SQL行列转换
-- http://www.programbbs.com/doc/4885.htm /* 标题:普通行列转换(version 2.0) 说明:普通行列转换(version 1.0)仅针对sql ser ...
- poj 1001 字符串乘法以及处理精度问题
#include<iostream> #include<cstring> using namespace std; int main() { string r; int n,d ...
- Vue中的scoped和scoped穿透
1.什么是scoped 在Vue文件中的style标签上有一个特殊的属性,scoped.当一个style标签拥有scoped属性时候,它的css样式只能用于当前的Vue组件,可以使组件的样式不相互污染 ...
- vue axios 跨域
qs是一个npm仓库所管理的包,可通过npm install qs命令进行安装. 1. qs.parse()将URL解析成对象的形式 2. qs.stringify()将对象 序列化成URL的形式,以 ...
- Java基础05-运算符
1.赋值运算符:= 2.算术运算符: (1)一元运算符: i++;先使用后自身加1 int i=1;int b=i+1; b=1 i=2 ++i;先对自身加1后使用 int j=1;int b=+ ...
- GridLayout(网格布局)
常用属性: 排列对齐: ①设置组件的排列方式: android:orientation="" vertical(竖直,默认)或者horizontal(水平) ②设置组件的 ...
- Linux网卡操作
单个网卡操作 [root@localhost ~]# ifdown eth0 #关闭网络 [root@localhost ~]# ifup eth0 #启动网络 网络服务: [root@localho ...
- DEDE利用Ajax实现调用当前登录会员的信息简要说明
其实这个功能在dede默认的模板上就有,只能算是在原有的功能上进行改造而已. 1.首先需要加载一个ajax的js文件进来 <script language="javascript&qu ...
- netty之==线程模型
1.1 netty线程模型本质遵循了Reactor的基础线程模型,所以得先介绍Reactor模型 1.2 Reactor模型 无论是C++还是Java编写的网络框架,大多数都是基于Reactor模 ...