Author Archives: wudi

199 Binary Tree Right Side View

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Total Accepted: 2068 Total Submissions: 7422

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

You should return [1, 3, 4].

 

[code]

binarySideView

[/code]

PsExec remote Invoke Command

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Without Configuration, PsExec could not execute successfully.

copy the following text and saved into .reg file in target machine:

Windows Registry Editor Version 5.00

[HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Policies\System]
“LocalAccountTokenFilterPolicy”=dword:00000001

 

Then PsExec could run successfully.

Repeated DNA Sequences

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Repeated DNA Sequences Total Accepted: 8816 Total Submissions: 47451 My Submissions Question Solution
All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: “ACGAATTCCG”. When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.

Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

For example,

Given s = “AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT”,

Return:
[“AAAAACCCCC”, “CCCCCAAAAA”].

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import java.util.*;
public class RepeatedDNASequences {
    public static List<String> findRepeatedDnaSequences(String s) {
 
        ArrayList<String> result = new ArrayList<String>();
 
        if(s== null || s.length() == 0) return result;
 
        HashMap<Character,Integer> map = new HashMap<Character,Integer>();
        map.put('A',0);
        map.put('C',1);
        map.put('G',2);
        map.put('T',3);
 
        HashSet<Integer> hashSet = new HashSet<Integer>();
        Set<Integer> unique = new HashSet<Integer>();
 
        int temp = 0;
        for(int i=0;i<s.length();i++){
            /*temp = 0;
              for(int j=0;j<10;j++){
              temp = (temp<<2) + map.get(s.charAt(i+j));
              }*/
            if(i<9){
                temp = (temp<<2) + map.get(s.charAt(i));
                // System.out.println(temp);
            }
            else{
                temp = (temp<<2) + map.get(s.charAt(i));
                temp &amp;= (1<<20) - 1;
                System.out.println(temp);
                if(hashSet.contains(temp) &amp;&amp; !unique.contains(temp)){
                    result.add(s.substring(i-9,i+1));
                    unique.add(temp);
                }
                else {
                    hashSet.add(temp);
                }
            }
        }
 
        return result;
    }
 
    public static void main(String[] args) {
        String s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT";
        System.out.println(findRepeatedDnaSequences(s));
    }
}

xcode 6.1 iphone simulator screen solution

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At the beginning, iphone solution is too big to suite small screen.

so I can not see all button and image.

the way to get around it, run simulator. from menu on the top of monitor choose “Windows” -> “Scale” -> 50%

or Use ShortCut, CMD+1, CMD+2,CMD+3

although now simulator screen looks much smaller but at lease I could see all button and image for UI

LeetCode OJ: Length of Last Word

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Length of Last Word

Total Accepted: 29154 Total Submissions: 100944

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example,
Given s = "Hello World",
return 5.

https://oj.leetcode.com/problems/length-of-last-word/

Solution: see the comment in the code , writed in Java

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public static int lengthOfLastWord(String s) {
    if(s.length() == 0) return 0; // input may be ""
 
    int result = -1;
 
    int skipEmpty = 0;
    // skip empty string at the end of input string s
    for(int i= s.length() - 1;i>=0;i--)
        {
            if(s.charAt(i) == ' ') continue;
            else {
                skipEmpty = i;
                break;
            }
        }
    //Start to Calculate
    for(int i= skipEmpty;i>=0;i--)
        {
            if(s.charAt(i) != ' ') continue;
            else {
                result = i;
                break;
            }
 
        }
 
    return skipEmpty - result;
}

emacs completion – xrefactory

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By default, Emacs just provide some simple auto complete. For example, Alt + /, it will invoke ‘dabbrev-expand’.
and ‘dabbrev-expand’ is to find and complete what you enter once it find match words in all ‘buffer’ opened.

xrefactory is excellent commerce plugin for emacs, xrefactory is free for C/Java.

Here is how to enable xrefactory in emacs.

xref-make

http://www.xref.sk/xrefactory/downloads/1.6.10/xrefactory-1.6.10-src.tgz

uncompress the tgz file，tar zxvf xrefactory-1.6.10-src.tgz

go into folder ‘xref-any’ , type command ‘make’, it will compile and finish to get ‘xref’

In .emacs file, add following code (replace ‘path/to’ to the path you want)

(add-to-list ‘load-path “/path/to/xref-any/env/emacs”)
(add-to-list ‘exec-path “/path/to/xref-any/src”)
(load “xrefactory”)
;;(load “/path/to/env/emacs/xrefactory.el”)

run ‘M-x eval-current-buffer’ , Now you could use xrefactory , Press ‘F8’ key to call ‘xref-completion’

xref is very powerfull, but we still has some other tools such as ‘gccsense, gtags’, I will add these two later on.

 

preprocess_new () at tccpp.c:3199 tcc的预处理

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// 这些函数算是比较基本的了，看看就明白

static inline int is_space(int ch)
{
return ch == ‘ ‘ || ch == ‘\t’ || ch == ‘\v’ || ch == ‘\f’ || ch == ‘\r’;
}
static inline int isid(int c)
{
return (c >= ‘a’ && c <= ‘z’) || (c >= ‘A’ && c <= ‘Z’) || c == ‘_’;
}
static inline int isnum(int c)
{
return c >= ‘0’ && c <= ‘9’;
}
static inline int isoct(int c)
{
return c >= ‘0’ && c <= ‘7’;
}
static inline int toup(int c)
{
return (c >= ‘a’ && c <= ‘z’) ? c – ‘a’ + ‘A’ : c;
}

preprocess_new () at tccpp.c:3199

初始化isidnum表

/* init isid table */
for(i=CH_EOF;i<256;i++)
isidnum_table[i-CH_EOF] = isid(i) || isnum(i);

// 该函数剩下几行把tcc的关键字插入 ptable (代码如下）

[code]

static TokenSym *tok_alloc_new(TokenSym **pts, const char *str, int len)
{
TokenSym *ts, **ptable;
[/code]

[code]

//在头文件tcctok.h定义了tcc的关键字

static const char tcc_keywords[] =
#define DEF(id, str) str “\0”
#include “tcctok.h”
#undef DEF
;
[/code]

ST_FUNC TokenSym *tok_alloc(const char *str, int len)