努橙刷题编

https://leetcode.com/problems/reverse-bits/

最简单的想法是位运算一位一位的挪，用for循环搞定（要注意优先级和结合性的问题）：

然后看到了别人一个不用循环的解法：
https://leetcode.com/problems/reverse-bits/discuss/54741/O(1)-bit-operation-C++-solution-(8ms)
很巧妙的利用了交换的时候这种特性：abcdefgh -> efghabcd -> ghefcdab -> hgfedcba 好困啊，只好用简单题交差啦～(￣▽￣～)(～￣▽￣)～

2018年12月30日星期日

[LeetCode] 5. Longest Palindromic Substring

https://leetcode.com/problems/longest-palindromic-substring/

解法一：使用了在水中的鱼的博客看到的第一种思路（http://fisherlei.blogspot.com/2012/12/leetcode-longest-palindromic-substring.html），对于任意一个index，分别向前、向后查找可能的palindrome。因为palindrome有“aba”和“abba”两种形式，所以还需要照顾到当s[index]==s[index + 1]这种情况。时间复杂度 O(n^2)。代码如下：

明天再想想其他解法 \(▔▽▔)/

2018年4月2日星期一

Bash 的基本语法规则

1. 函数传参

#!/bin/bash

test_function() {

echo $1 $2

}

test_function "hello" "world"

Result:

~$ ./example.sh

hello world

2. 分割字符串

#!/bin/bash

LINE="test_name,test_address,test_phone_number"

name=$(echo $LINE | cut -d ',' -f 1)

address=$(echo $LINE | cut -d ',' -f 2)

phone_number=$(echo $LINE | cut -d ',' -f 3)

echo $name

echo $address

echo $phone_number

Result:

~$ ./example.sh

test_name

test_address

test_phone_number

3. 读取文件

#!/bin/bash

i=0

SPACE=" "

while read line

echo $i $SPACE $line

i=$((i+1))

done < test.txt

Result:

~$ cat test.txt

hahaha

haha

~$ ./example.sh

0 hahaha

1 haha

2 ha

4. 正则表达式匹配

#!/bin/bash

LINE="This is a test string. Let's see if the UUID will be matched: ea77b80d-d17f-476a-b55f-916423d02dcc. Hahaha!"

uuid=$(echo $LINE | grep -E -o '[0-9a-f]{8}-[0-9a-f]{4}-[1-5][0-9a-f]{3}-[89ab][0-9a-f]{3}-[0-9a-f]{12}')

echo $uuid

Result:

~$ ./example.sh

ea77b80d-d17f-476a-b55f-916423d02dcc

2018年4月1日星期日

Hazelcast 学习笔记

(1) Install maven - https://stackoverflow.com/a/17481557

~$ ls /Library/Java/JavaVirtualMachines/jdk1.8.0_161.jdk/Contents/Home

export JAVA_HOME=/Library/Java/JavaVirtualMachines/jdk1.8.0_161.jdk/Contents/Home

(2) build a maven package - https://www.mkyong.com/maven/how-to-build-project-with-maven/

(3) Exception in thread "main" com.hazelcast.license.exception.InvalidLicenseException: License Key not configured!

http://docs.hazelcast.org/docs/latest-development/manual/html/Getting_Started/Setting_The_License_Keys.html

(4) Replication vs. partitioning

https://www.youtube.com/watch?v=W1NmbcuomVM

(5) Most common use case: being a cache on top of a database

(6) Hazelcast has two modes: embedded mode and client-server mode

2017年11月29日星期三

About memory leak

(WIP)
* jvisualvm
** ssh -X
* jmap
* example of memory leak: https://dzone.com/articles/memory-leak-andjava-code

2017年9月27日星期三

Trie Implementation

 public class Trie {  
      static final int ALPHABET_SIZE = 26;  
      static class TrieNode {  
           TrieNode[] children = new TrieNode[ALPHABET_SIZE];  
           boolean isEndOfWord;  
           TrieNode() {  
                isEndOfWord = false;  
                for (int i = 0; i < ALPHABET_SIZE; i++) {  
                     children[i] = null;  
                }  
           }  
      }  
      static TrieNode root;  
      // if not parent, inserts key into trie  
      // if the key is prefix of trie node, just mark leaf node  
      static void insert(String key) {  
           int level;  
           int length = key.length();  
           int index;  
           TrieNode pCrawl = root;  
           for (level = 0; level < length; level++) {  
                index = key.charAt(level) - 'a';  
                if (pCrawl.children[index] == null) {  
                     pCrawl.children[index] = new TrieNode();  
                }  
                pCrawl = pCrawl.children[index];  
           }  
           // mark the last node as leaf  
           pCrawl.isEndOfWord = true;  
      }  
      // Returns true if key presnts in trie, else false  
      static boolean search(String key) {  
           int level;  
           int length = key.length();  
           int index;  
           TrieNode pCrawl = root;  
           for (level = 0; level < length; level++) {  
                index = key.charAt(level) - 'a';  
                if (pCrawl.children[index] == null) {  
                     return false;  
                }  
                pCrawl = pCrawl.children[index];  
           }  
           return pCrawl != null && pCrawl.isEndOfWord;  
      }   
 }

2016年10月8日星期六

[LeetCode] #200 Number of Islands

public class Solution {
    public int numIslands(char[][] grid) {
        int result = 0;
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return result;
        }
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1') {
                    result++;
                    dfs(grid, i, j);
                }
            }
        }
        return result;
    }
    
    private void dfs(char[][] grid, int i, int j) {
        if (grid[i][j] != '1') {
            return;
        }
        
        grid[i][j] = '2';
        if (i - 1 >= 0) {
            dfs(grid, i - 1, j);
        }
        if (j - 1 >= 0) {
            dfs(grid, i, j - 1);
        }
        if (i + 1 < grid.length) {
            dfs(grid, i + 1, j);
        }
        if (j + 1 < grid[0].length) {
            dfs(grid, i, j + 1);
        }
    }
}

订阅：博文 (Atom)