m*********a
发帖数: 3299 | 1 node dummy(0);
node *ptr
node * &curr=ptr;
其实是node *是个数据类型(ptr的数据类型）,curr是ptr的另一个名字(&curr=ptr)，
ptr和curr的内存地址一样。
这和
int x;
int &y=x;是一样的，y是int这个数据类型(x的数据类型），y是x的另一个名字(&y=x)
，地址一样。 |
|
i****n
发帖数: 42 | 2 请教一个 C++ linked list/Tree related 的小问题:
struct Node
{
int data;
Node *next;
}
Compared against Node *curr, what does Node *&curr or Node **curr exactly
mean? Thanks a lot. |
|
y***n
发帖数: 1594 | 3 Node *curr: pointer to a node.
Node *&curr: reference to pointer to a node.
Node ** curr : pointer to pointer to a node.
每个都写一下，就会加深理解。。
还有这些问题去StackOverflow, 很多牛人会来回答，还可以得分 |
|
I****p
发帖数: 101 | 4 好像 Curr Biol 对题材非常 selective，但 IF 只比 PNAS 高一点。在遗传、细胞、
发育领域，对 Job Market 来说，Curr Biol 好多少？谢谢！ |
|
b******s
发帖数: 1089 | 5 明年就回国了，现在想把手头的文章发掉。老板坚持要nature communications, 可能
是因为我们curr biol和PNAS都发过。
我本来也没特别倾向，不过才发现，按照中科院分区nature communications是2区杂志
，虽然影响因子过10，但是分区里和Plos One都一个区的。。。
跑来问问回国用哪个杂志性价比最好？我自己觉得curr biol速度比较快，相对也友好
，PNAS慢一点，但也还可以。不知道Nature communications是不是跟其他nature子刊
一样时间会拖得臭长？ |
|
s******y
发帖数: 28562 | 6 NC是会拖得很长，审查起来其实挺严的。
我和前老板手头有一篇就是投那里去的，除了新意可以要求稍微低一点之外，对数据质
量和数量要求挺高的。一年了，我们还在补数据。
如果Curr Biol 对你们友好的话你应该先发这个。你可以对老板这么说：
１。很多你的朋友在投稿中都发现Nature Comm 审稿时间拖得很长，如果投这个的话有
可能你离开了还没有发表，这样就会需要别人来帮你收尾，将是一件很麻烦的事情。
２。Curr Biol 是一个老牌杂志，读者比Nature Comm只会多不会少。 |
|
j**d
发帖数: 825 | 7 【 以下文字转载自 Biology 讨论区 】
发信人: jsad (沙沙), 信区: Biology
标 题: 求文献：Curr Opin Rheumatol. 2011 May;23(3):278-81.
发信站: BBS 未名空间站 (Thu Apr 7 09:39:40 2011, 美东)
Curr Opin Rheumatol. 2011 May;23(3):278-81.
Re-evaluation of antimalarials in treating rheumatic diseases: re-
appreciation and insights into new mechanisms of action.
请发到j*****[email protected]
多谢多谢。 |
|
w**********s
发帖数: 72 | 8 BACE1 structure and function in health and Alzheimer's disease.
Curr Alzheimer Res. 2008 Apr;5(2):100-20.
Authors: Cole SL, Vassar R.
My email is w**********[email protected]
Thansk very much! |
|
e******n
发帖数: 193 | 9 Curr Opin Infect Dis. 2009 Feb;22(1):87-91.
Neisseria gonorrhoeae and emerging resistance to extended spectrum
cephalosporins.
Tapsall JW.
World Health Collaborating Centre for STDs, Department of Microbiology, The
Prince of Wales Hospital, Sydney, Australia. j*******[email protected]
PMID: 19532086 [PubMed - indexed for MEDLINE]
Thank you in advance
email: c***********[email protected] |
|
l***7
发帖数: 50 | 10 1. Curr Opin Hematol. 2010 Jul;17(4):271-5.
Reprogramming adult hematopoietic cells.
Kaneko S, Otsu M, Nakauchi H.
2. Cytotherapy. 2009;11(8):980-9.
Application of induced pluripotent stem cells to hematologic disease.
Kim PG, Daley GQ.
Please send to y*****[email protected]
Thank you very much. |
|
j**d
发帖数: 825 | 11 Curr Opin Rheumatol. 2011 May;23(3):278-81.
Re-evaluation of antimalarials in treating rheumatic diseases: re-
appreciation and insights into new mechanisms of action.
请发到j*****[email protected]
多谢多谢。 |
|
g******0
发帖数: 290 | 12 Curr Drug Targets. 2011 Apr 1;12(4):563-78.
Dasatinib: An Anti-Tumour Agent via Src Inhibition.
Gnoni A, Marech I, Silvestris N, Vacca A, Lorusso V.
Thank you in advance. |
|
z*******6
发帖数: 679 | 13 Curr Mol Med. 2008 Aug;8(5):384-92.
Interleukin-4, interleukin-13, signal transducer and activator of
transcription factor 6, and allergic asthma.
Kuperman DA, Schleimer RP.
n******[email protected]
Thank you so much. |
|
|
|
|
|
|
L****e
发帖数: 499 | 19 Curr Opin Lipidol. 2014 Feb 22.
Comparative gene identification-58/α/β hydrolase domain 5: more than just
an adipose triglyceride lipase activator? |
|
m*******8
发帖数: 256 | 20 我感觉国内认的顺序是：PNAS, Nat Comm, Curr Biol. |
|
A**I
发帖数: 73 | 21 感觉国内顺序是NC，PNAS，Curr Biol. |
|
C*********m
发帖数: 213 | 22 跪谢！请邮 [email protected]/* */
作者:
Marinangeli C, Didier S, Vingtdeux V
文题:
AMPK in Neurodegenerative Diseases: Implications and Therapeutic
Perspectives.
期刊名:
Curr Drug Targets
期刊年份:
2016
卷(期),起止页码:
17(8):890-907
Doi:
10.2174/1573399811666150615150235
全文链接:
http://www.eurekaselect.com/132222/article
数据库名称:
Bentham Science Publishers |
|
s********l
发帖数: 1195 | 23 Curr Opin Lipidol. 2010 Jan 30. [Epub ahead of print]
Strategies for identifying the genetic basis of dyslipidemia: genome-wide
association studies vs. the resequencing of extremes.
Khor CC, Goh DL.
w****[email protected]
Thanks! |
|
A*r
发帖数: 2253 | 24 Somatostatin receptor-targeted anti-cancer therapy.
Sun LC, Coy DH.
Curr Drug Deliv. 2011 Jan 1;8(1):2-10.
对我很重要的文章，发包子给先帮我找到这篇文章的人。
Email to: g**[email protected]
Thanks. |
|
s*****a
发帖数: 7 | 25 Discovery of Notch-Sparing gamma-Secretase Inhibitors.
Curr Alzheimer Res. 2010 Jan 21. [Epub ahead of print]
Thanks and bow
My email: w******[email protected] |
|
p********n
发帖数: 165 | 26 Space O(1)
Time O(n)
// Given a string, return the longest substring that contains at most // two
characters.
// solution: scan the string and update the first and second letter's last
occurrence
// indices, and update the solution's start index.
int FindLength(const string& input) {
if (input.size() <= 2) {
return input.size();
}
int solu_start = 0;
int first_end, second_end;
char first = input[0];
char second;
int curr = 1;
while (curr < input.size() && in... 阅读全帖 |
|
p********n
发帖数: 165 | 27 Space O(1)
Time O(n)
// Given a string, return the longest substring that contains at most // two
characters.
// solution: scan the string and update the first and second letter's last
occurrence
// indices, and update the solution's start index.
int FindLength(const string& input) {
if (input.size() <= 2) {
return input.size();
}
int solu_start = 0;
int first_end, second_end;
char first = input[0];
char second;
int curr = 1;
while (curr < input.size() && in... 阅读全帖 |
|
y******8
发帖数: 1764 | 28 Abbreviated Journal Title ISSN {2012} Total Cites Impact Factor
5-Year Impact Factor Immediacy Index {2012} Articles Cited Half-
Life Eigenfactor Score Article Influence Score
CA-CANCER J CLIN 0007-9235 13722 153.459 88.55 27.04 25
3.3 0.0517 29.478
NEW ENGL J MED 0028-4793 245605 51.658 50.807 12.667 360
8 0.65776 21.494
REV MOD PHYS 0034-6861 35720 44.982 51.882 6.478 46 10
0.13048 32.634... 阅读全帖 |
|
a********k
发帖数: 2273 | 29 JCR Year and Edition: 2010 Science
Abbreviated Journal Title ISSN {2010} Total Cites Impact Factor
5-Year Impact Factor Immediacy Index {2010} Articles Cited Half-
Life Eigenfactor Score Article Influence Score
CA-CANCER J CLIN 0007-9235 9801 94.262 70.216 8.667 18
3.8 0.04923 24.782
ACTA CRYSTALLOGR A 0108-7673 13944 54.333 24.717 0.629 70
6.2 0.04... 阅读全帖 |
|
l*********8
发帖数: 4642 | 30 我写一个linked list flatten和restore吧。 可能还有bugs
Node * flatten(Node * curr, Node * tail = NULL, Node * parent = NULL) {
if (!curr) return tail;
if (tail)
tail->next = current;
tail = current;
Node * next = curr->next;
if (curr->child)
tail = flatten(curr->child, tail, next ? curr : parent);
else if (!next)
curr->child = parent;
return flatten(next, tail, parent);
}
void restore(Node * curr) {
while (curr) {
Node * next = curr->next;
... 阅读全帖 |
|
i**********e
发帖数: 1145 | 31 这是我的代码，虽然没有那么漂亮简洁，但是比较容易懂。
我可以尝试把代码缩短，但是缩短之后就很难理解了。
void post_order_traversal_iterative(BinaryTree* root) {
bool down = true;
BinaryTree *prev;
stack s;
s.push(root);
while (!s.empty()) {
BinaryTree *curr = s.top();
if (down) {
if (curr->left) {
s.push(curr->left);
} else {
if (curr->right) {
s.push(curr->right);
} else {
down = false;
cout << curr->data << " ";
s.pop();
prev = cu... 阅读全帖 |
|
c*****m
发帖数: 315 | 32 搞了一上午搞出来了第一题，包含以下几点
1. 基本想法类似于HEAPSORT， 每次取HEAP 的最小值，和BST tree 当前最小位置的值
交换，然后再维护HEAP 的结构，同时取BST 中序遍历的下一个值。在整个过程中，树
的一部分变成了BST TREE, 另一部分还是HEAP 结构。
下面介绍相关的步骤。
2. 取HEAP 的最小值：粗略的想法是用HEAP[0]。这里有个问题：当HEAP[0] 被并入BST
部分以后，最小值不再是HEAP[0]，而是HEAP[0] 的右子树。可以证明HEAP[0] 被并入
BST 的时候，它的左子树已经全部在BST 里了，而它的右子树都不在BST里，因此可以
TRACK 当前HEAP 的ROOT。
3.维护HEAP 的结构。算法和SIFTDOWN 一样，但要注意的是SIFT DOWN 的时候必须要检
查当前节点的子节点是不是在BST 部分，如果是，就停止交换。要做到这一点，只需要
TRACK 住当前HEAP 的最小值，BST 里的元素都会小于它。
代码如下（HEAP 用数组heap 表示）：
//取一个BST子树最小节点的索引， 辅助函数
... 阅读全帖 |
|
e*******i
发帖数: 56 | 33 No map is needed. Source is as following:
///////////////
#include
using namespace std;
int findLongest(char *s, char **start)
{
if(!s) return 0;
char first, second;
char *beginOfPreviousChar;
int currMax=0;
char *curr=s;
first=*s;
*start=s;
while(*curr)
{
if(*curr!=first) break;
currMax++;
curr++;
}
if( !*curr ) return currMax; //end of string
else
{
second=*curr;
beginOfPreviousChar=curr... 阅读全帖 |
|
l*****s
发帖数: 774 | 34 iterative
class Solution {
public:
void connect(TreeLinkNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
TreeLinkNode *curr = root;
while(curr)
{
TreeLinkNode *left_most = NULL;
TreeLinkNode *prev = NULL;
while(curr)
{
if(!left_most) left_most = curr->left?curr->
l... 阅读全帖 |
|
a*****y
发帖数: 22 | 35 假设Quack的顺序是递增
我的思路是，想办法找到最大的元素（尾），然后push进一个标志（如MIN_INT）;下次
peek的时候，只pop非MIN_INT的元素，直到再次遇到最大的那个元素，此时可以pop标
志，直至空
void PopQuackToArray(Quack& quack, vector& array) {
vector tmp;
int last = quack.pop();
tmp.push_back(last);
int curr;
while (!quack.empty()) {
curr = quack.pop();
if (curr != last) {
break;
}
tmp.push_back(last);
}
if (quack.empty()) {
if (curr < last) {
array.push_back(curr);
... 阅读全帖 |
|
a*****y
发帖数: 22 | 36 假设Quack的顺序是递增
我的思路是，想办法找到最大的元素（尾），然后push进一个标志（如MIN_INT）;下次
peek的时候，只pop非MIN_INT的元素，直到再次遇到最大的那个元素，此时可以pop标
志，直至空
void PopQuackToArray(Quack& quack, vector& array) {
vector tmp;
int last = quack.pop();
tmp.push_back(last);
int curr;
while (!quack.empty()) {
curr = quack.pop();
if (curr != last) {
break;
}
tmp.push_back(last);
}
if (quack.empty()) {
if (curr < last) {
array.push_back(curr);
... 阅读全帖 |
|
m**********e
发帖数: 22 | 37 我来贡献一个用queue的方法。C#写的：
// Iterative approach
public List generateParenthesisIter(int n)
{
Queue queue = new Queue(
);
queue.Enqueue(new ParenthesisWrapper("",0,0));
List result = new List();
while (queue.Count > 0)
{
ParenthesisWrapper curr = queue.Dequeue();
if (curr.left == n)
{
if ... 阅读全帖 |
|
X******g
发帖数: 10 | 38 第二题用Morris Inorder遍历就行。
vector inorderTraversal(TreeNode* root)
{
vector res;
TreeNode* curr=root;
while(curr)
{
if(!curr->left)
{
res.push_back(curr->val);
curr=curr->right;
}
else
{
TreeNode* lrMost=curr->left;
while(lrMost->right&&lrMost->right!=curr)
lrMost=lrMost->right;
if(!lrMost->r... 阅读全帖 |
|
p*i
发帖数: 411 | 39 int myatoi(const string& str) {
const int n = str.size();
int curr = 0; // current index
while ((curr < n) && (isspace(str[curr]))) curr++; // skip leading
spaces
int sign = 1; // positive
if (str[curr] == '-') {
sign = -1; curr++;
}
if (curr == n)
throw runtime_error("Invalid input");
// prepare overflow check
int base, extra;
if (sign == 1) {
base = numeric_limits::max() / 10;
extra = numeric_limits::max() % 10;
... 阅读全帖 |
|
l*n
发帖数: 529 | 40 来自主题: JobHunting版 - G新鲜面经 仔细考虑了下，1.2这么做比较简单：就是把数组遍历一遍，但是每次要调头，如果调
头的时候没有候选了就fail.
写了个java的code
[1,2,4,5,6]给出的所有结果是
[1, 4, 2, 6, 5]
[1, 5, 4, 6, 2]
[1, 5, 2, 6, 4]
[1, 6, 4, 5, 2]
[1, 6, 2, 5, 4]
[2, 4, 1, 6, 5]
[2, 5, 4, 6, 1]
[2, 5, 1, 6, 4]
[2, 6, 4, 5, 1]
[2, 6, 1, 5, 4]
[4, 5, 2, 6, 1]
[4, 5, 1, 6, 2]
[4, 6, 2, 5, 1]
[4, 6, 1, 5, 2]
[5, 6, 2, 4, 1]
[5, 6, 1, 4, 2]
List> reorder(int[] arr) {
assert (arr != null);
Arrays.sort(arr);
List> ol = new ArrayList>();
... 阅读全帖 |
|
z****h
发帖数: 164 | 41 public ListNode deleteDuplicates(ListNode head) {
// Start typing your Java solution below
// DO NOT write main() function
if(head == null || head.next == null) return head;
ListNode curr = head;
ListNode pre = null;
boolean isdup = false;
while(curr.next != null)
{
if(curr.val == curr.next.val){
isdup = true;
curr.next = curr.next.next;
}else
{
... 阅读全帖 |
|
r*******n
发帖数: 3020 | 42 这个题有点意思。
ListNode* insert(ListNode* head){
if(head==NULL)
return;
ListNode* curr=head->next;
ListNode* dummyHead= new ListNode(0);
dummyHead->next=head;
while(curr!=NULL){
ListNode* pre=dummyHead;
// Find where to insert
while(p!=curr){
if(p->val>=curr->val)
break;
pre=p;
p=p->next;
}
if(p==curr){
//curr's value is largest so far
curr=curr->next;
}else{
//d... 阅读全帖 |
|
m*****f
发帖数: 1243 | 43
1. curr = prev = root
2. Initialize stack
3. while (true)
4. if ( curr is prev or a child of prev) //Traversing downwards
5. if (curr is a parent) //More traversing to do
6. stack.push(curr)
7. prev = curr
8. curr = curr.left
9. else //curr is an orphan: go upward
10. print (curr)
11. prev = curr
12. curr = stack(top)
13. if ( curr.left = prev) //Traversing upwa |
|
i**********e
发帖数: 1145 | 44 再试试这个更简洁的思路，去除了使用down的变量。
void post_order_traversal_iterative3(BinaryTree* root) {
stack s;
s.push(root);
BinaryTree *prev = NULL;
while (!s.empty()) {
BinaryTree *curr = s.top();
if (curr->left == prev) {
if (curr->right)
s.push(curr->right);
} else if (!prev || prev->left == curr || prev->right == curr) {
if (curr->left)
s.push(curr->left);
else if (curr->right)
s.push(curr->right);
} else {
cout << curr->data... 阅读全帖 |
|
r******n
发帖数: 170 | 45 感觉可以模仿postorder的iterative 方法写，也应该得加个visited flag
如下：
void mirror(Node* root)
{
if (root == NULL)
return;
mirror(root->left);
mirror(root->right);
Node* tmp =root->left;
root->left=root->right;
root->right=tmp;
}
void mirror_iter(Node* root)
{
if (root == NULL)
return;
stack nodeStack;
nodeStack.push(root);
while (!nodeStack.empty())
{
Node* curr=nodeStack.top();
if (curr->left != NULL && curr-... 阅读全帖 |
|
t**********h
发帖数: 2273 | 46 public static void main(String[] args){
Node n1 = new Node(1);
Node n2 = new Node(2);
Node n3 = new Node(3);
Node n4 = new Node(3);
Node n5 = new Node(3);
Node n6 = new Node(4);
Node n7 = new Node(4);
Node n8 = new Node(5);
n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = n5;
n5.next = n6;
n6.next = n7;
n7.next = n8;
Node result = remove(n... 阅读全帖 |
|
l****i
发帖数: 230 | 47 贴一下自己的code
BTW，我虽然是CS的PHD但自己本科硕士都不是CS，所以coding很烂，不是什么大牛，跟
版上很多人比差距不小，说一下这个也权当是对大家的鼓励吧
int nQueens(int n){
if(n<=0) return 0;
int *placement = new int[n];
assert(placement);
int cnt = nQueensHelper(placement, n, 0);
delete[] placement;
return cnt;
}
int nQueensHelper(int *placement, int sz, int curr){
if(curr>=sz) return 1;
int cnt = 0;
for(int i=0; i
placement[curr] = i;
if(validatePlacement(placement, sz, curr))
cnt += nQueensHelper(placement, sz, c... 阅读全帖 |
|
p*****2
发帖数: 21240 | 48
都差不多吧。
public ListNode deleteDuplicates(ListNode head)
{
ListNode curr=head;
ListNode tail=head=null;
for(ListNode prev=null;curr!=null;prev=curr,curr=curr.next)
{
if((prev==null || prev.val!=curr.val) && (curr.next==null ||
curr.val!=curr.next.val))
{
if(tail==null)
{
tail=curr;
head=curr;
}
else
{
... 阅读全帖 |
|
p*****2
发帖数: 21240 | 49 我写了一个，有点繁琐，还没测试。
void connectSibling(Node root)
{
assert(root!=null);
Node curr=root;
while(curr!=null)
{
Node prev=null;
Node next=null;
while(curr!=null)
{
if(curr.left!=null)
{
if(prev==null)
{
prev=curr.left;
next=prev;
}
... 阅读全帖 |
|
p*****2
发帖数: 21240 | 50
瞎写了一个。
final int SIZE=60*60; //seconds in one hour
int[] arr=new int[SIZE];
int hour=0;
int minute=0;
long last=0;
long currSecond(){
return System.currentTimeMillis()/1000;
}
long clear(){
long curr=currSecond();
if(curr>last){
if(curr-last>=SIZE){
hour=0;
minute=0;
Arrays.fill(arr,0);
}
else{
if(curr-last>=60){
... 阅读全帖 |
|
