etihad 103

_usf_file If ["$?"-eq 0]; Then echo "[] set Newcopyright statement in $deploy _usf_file OK." Fi Else echo "[Ten] Error:set newcopyright statement in $deploy _usf_file fail." Fi Avoid replacing the text with a separator character/, using # as the separator. To run a script that prompts for an error: Sed:-E expression #1, char 103:unknown option to ' s ' Check the script found that similar three-line replacement operation, the fir

ORA-00600: internal error code, arguments: [kcblasm_1], [103], [], [], [], [], [], [], [], ora-00600kcblasm_1 This error has been shown in the alarm log for the past two days: ORA-00600: internal error code, arguments: [kcblasm_1], [103], [], [], [], [], [], [], [] Tue Aug 12 09:20:17 CST 2014 Errors in file/u01/app/oracle/admin/orcl/udump/orcl_ora_29974.trc: ORA-00600: internal error code, arguments: [kcbl

This error has been shown in the alarm log for the past two days: ORA-00600: Internal error code, arguments: [kcblasm_1], [103], [], [], [], [], [], [], [] Tue Aug 12 09:20:17 CST 2014 Errors in file/u01/APP/Oracle/admin/orcl/udump/orcl_ora_29974.trc: ORA-00600: Internal error code, arguments: [kcblasm_1], [103], [], [], [], [], [], [], [] Tue Aug 12 09:30:17 CST 2014 Errors in file/u01/APP/Oracle/admin/orc

[LeetCode-interview algorithm classic-Java implementation] [103-Binary Tree Zigzag Level Order Traversal (Binary Tree layered Z-shaped Traversal)], zigzagleetcode [103-Binary Tree Zigzag Level Order Traversal (Binary Tree layered Z-shaped Traversal )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question Given a binary tree, return the zigzag level

UVA 103 Stacking BoxesBackgroundSome Concepts in Mathematics and computer science is simple in one or both dimensions but become more complex when Extende D to arbitrary dimensions. Consider solving differential equations in several dimensions and analyzing the topology of a n-dimensional hypercube. The former is much more complicated than it one dimensional relative while the latter bears a remarkable resemblance to I TS "Lower-class" cousin.The prob

Topic Portal1 /*2 Shortest way: Not only scan the edges, but also scan the points; There are two cases, one is just at the midpoint, that is, from the u,v, then the last/23 there is a different from the u,v, both of the distance is L4 The template was wrong and it was a long time: (5 */6#include 7#include 8#include 9#include Ten#include One#include A using namespacestd; - - Const intMAXN = 1e5 +Ten; the Const intINF =0x3f3f3f3f; - intD[MAXN]; - intCNT[MAXN]; - BOOLVIS[MAXN]; +vectorint,int> >

C ++ Primer study note _ 103 _ special tools and technology, primer_103Special tools and technologies-class member pointers The member pointer can be used to obtain the pointer of a specific member, and then obtain the member from an object or another object. The member pointer should contain the class type and the member type. I. Declare a member pointer Test class: class Screen{public: typedef std::string::size_type index; char get() const;

UV 103 Stacking Boxes (DP), stackingboxesBackground Some concepts in Mathematics and Computer Science are simple in one or two dimensions but become more complex when extended to arbitrary dimensions. Consider solving differential equations in several dimensions and analyzing the topologyN-Dimen1_hypercube. The former is much more complicated than its one dimen1_relative while the latter bears a remarkable resesponance to its ''lower-class'' cousin. T

Python one-day training 103 ---- Django template exercises, pythondjango Requirements Request. META is a Python dictionary that contains all the Header information of this HTTP request, such as the user IP address and user Agent. You can obtain this dictionary through request. META. items. You must output the META information to the webpage. Output result Source code index.html Views. py from django.shortcuts import render# Create your views here.d

#include 　　Memory Search UVA 103

*temp; - intLev =0; + while(!Q.empty ()) { A intSize =q.size (); at while(size--) { -temp =Q.front (); - Q.pop (); -Row.push_back (temp->val); - if(Temp->left! =NULL) { -Q.push (temp->Left ); in } - if(Temp->right! =NULL) { toQ.push (temp->Right ); + } - } the if(Lev%2) { * intn =row.size (); $ for(inti =0; I 2; i++) {Panax Notoginsen

30 1023 15 7 9 11 340 50 34 24 14 49 10 11 12 13 1431 4 18 8 27 1744 32 13 19 41 191 2 3 4 5 680 37 47 18 21 9Sample Output53 1 2 4 547 2 5 6Test instructionsFor n-dimensional graphics, their side length is {D1,D2,D3...DN}, and for two n-dimensional graphs, if all of the edge lengths of one of them correspond to an edge length that is less than the other in any order, then the lock can be nested to another. For example a{1,2}, b{2,3}, a all edge lengths correspond to the length of the edge less

Given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).For example:Given binary Tree {3,9,20,#,#,15,7}, 3 9 20 / 15 7Return its zigzag level order traversal as:[ [3], [20,9], [15,7]]Idea: Using the sequence traversal, through the queue, sequentially traverse each layer, if the number of layers is even, then the number of the layer is flipped, using the vector's r

The method described in the previous section only creates a static library and cannot be debugged because creating a static library project directly is not an executable file.The method that this article describes creates a static library breakpoint that works when you debug.In order to be able to debug a static library, it should be in an executable project (for example, single View application) and then add a static library inside it, adding the method as shown:Click on the project configurati

Given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).For example:Given binary Tree {3,9,20,#,#,15,7} , 3 / 9 / 7Return its zigzag level order traversal as:[ 3], [20,9], [15,7]]Train of thought: With two fork tree horizontal sequence problem-solving idea is similar, get the horizontal order result, then dual number chain list inversion can. ‘The code is as follows:/*

Label: lis Link: Ultraviolet A 103 Question: For n-dimensional images, their edge lengths are {D1, D2, D3.. DN}. For two n-dimensional images, find all the edge lengths that satisfy one of them. FollowAny orderBothOne-to-one correspondenceSmaller than the length of the other side, the number of such longest sequence, and print the path of any of the longest string, For example: A (,), B (,), C (,), A and B do not meet, but C and B meet Analysis:

) = Sumgreen (0) + DATAARRC (1)Sumblue (0) = Sumblue (0) + DATAARRC (0)Sumnum (0) = Sumnum (0) + 1 ' counter plus oneNextNext Step two: Calculate the blur value of the first row of pixels in addition to the other pixels in the 0,0 point, which takes full advantage of the first step of the information. The code for this step is detailed in the project. In the third step, based on the first line of the image, the fuzzy values of the following lines are calculated at once, there are some actions to

Systematic training, inspirational challenge Programming Competition-code sorting 103 ~ 134 [preliminary] In September 6, 2014, I got it here. I have finished the preliminary refresh and stopped it first. I read the directory in the intermediate article. I used to refresh it all before, but I don't have a system refresh. I want to refresh it several days later, let's take a look at the java interview book. It is expected to take 10 days to complete. R

Source of the topic:https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Test Instructions Analysis:The z-word width traverses the tree. Topic Ideas:This problem can be compared with the Trickery method. First get the result of the width traversal, and then flip the second layer to be able. Code (Python):　　#Definition for a binary tree node.#class TreeNode (object):#def __init__ (self, x):#self.val = x#self.left = None#self.right = Noneclasssolution (object):

Given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).For example:Given binary Tree {3,9,20,#,#,15,7} , 3 / 9 / 7Return its zigzag level order traversal as:[ 3], [20,9], [15,7]]Problem Solving Ideas:Add a reverse statement directly to the title and the Java implementation is as follows: Public listJava for Leetcode 103