b*n
2 楼
想上的可以搞这个,authorized dealer,价格也还成
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
r*5
4 楼
等崩到2k再说
【在 b******n 的大作中提到】
: 想上的可以搞这个,authorized dealer,价格也还成
: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
【在 b******n 的大作中提到】
: 想上的可以搞这个,authorized dealer,价格也还成
: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
d*l
6 楼
hotdigital 到底是和什么一家的
在canon list上面找不到他,但大家都说是authorized
【在 b******n 的大作中提到】
: 想上的可以搞这个,authorized dealer,价格也还成
: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
在canon list上面找不到他,但大家都说是authorized
【在 b******n 的大作中提到】
: 想上的可以搞这个,authorized dealer,价格也还成
: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
f*t
9 楼
http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=
【在 t**********h 的大作中提到】
: 完全不懂你在说什么啊
【在 t**********h 的大作中提到】
: 完全不懂你在说什么啊
t*e
10 楼
它家卖第一个就是 $2449.95,被俺拿下,后来它又涨到 $2499.95,再降回来。。。
【在 b******n 的大作中提到】
: 想上的可以搞这个,authorized dealer,价格也还成
: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
【在 b******n 的大作中提到】
: 想上的可以搞这个,authorized dealer,价格也还成
: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=300420450172
t*h
11 楼
多谢。
我只会150上的版本,赶紧学学这个版本
【在 f*******t 的大作中提到】
: http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=
我只会150上的版本,赶紧学学这个版本
【在 f*******t 的大作中提到】
: http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=
A*h
12 楼
又降啦!?
我忍,到五月初在说。
我忍,到五月初在说。
i*e
13 楼
面试不会让你写那么复杂算法的。
递归程序就只用五行,O(N) 复杂度。
http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
递归程序就只用五行,O(N) 复杂度。
http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
w*z
15 楼
这个好,看得懂,记得住。
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
H*r
17 楼
Very nice article!
那个5行的Bottom-up Approach (Worst case O(n) )
怎么感觉对degenerate tree仍然是O(N*N)?
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
那个5行的Bottom-up Approach (Worst case O(n) )
怎么感觉对degenerate tree仍然是O(N*N)?
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
p*y
18 楼
似乎确实是的,官方名字是“Focus Camera Inc.”,你看看这个帖子:http://m.slic
kdeals.net/forums/showthread.php?t=1945972
【在 d********l 的大作中提到】
: hotdigital 到底是和什么一家的
: 在canon list上面找不到他,但大家都说是authorized
kdeals.net/forums/showthread.php?t=1945972
【在 d********l 的大作中提到】
: hotdigital 到底是和什么一家的
: 在canon list上面找不到他,但大家都说是authorized
s*a
19 楼
no need for recursion. just trace back from each node to root, and print out
the first common ancester..(it should be log(N) for a binary tree ?)
the first common ancester..(it should be log(N) for a binary tree ?)
i*e
21 楼
You need to ask interviewer whether there's a parent node.
It makes a big difference in the algorithm.
If there's parent node, it can be improved to O(h) , h = height of tree.
out
【在 s*******a 的大作中提到】
: no need for recursion. just trace back from each node to root, and print out
: the first common ancester..(it should be log(N) for a binary tree ?)
It makes a big difference in the algorithm.
If there's parent node, it can be improved to O(h) , h = height of tree.
out
【在 s*******a 的大作中提到】
: no need for recursion. just trace back from each node to root, and print out
: the first common ancester..(it should be log(N) for a binary tree ?)
s*a
22 楼
but there is always parent node (except for the root) since it is a tree.
I guess you meant the implementation (e.g., array vs linked data structure
via pointers), but that should be irrelevant at the algorithmic level..
【在 i**********e 的大作中提到】
: You need to ask interviewer whether there's a parent node.
: It makes a big difference in the algorithm.
: If there's parent node, it can be improved to O(h) , h = height of tree.
:
: out
I guess you meant the implementation (e.g., array vs linked data structure
via pointers), but that should be irrelevant at the algorithmic level..
【在 i**********e 的大作中提到】
: You need to ask interviewer whether there's a parent node.
: It makes a big difference in the algorithm.
: If there's parent node, it can be improved to O(h) , h = height of tree.
:
: out
t*h
23 楼
我擦,这个好,帅
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
i*e
24 楼
for a REAL tree there's a root.
but in computer science, usually tree is defined as:
struct Node {
Node *left;
Node *right;
};
You cannot assume that each node has a parent node. Actually this is a great
question to ask your interviewer, because you're responsible for finding
out the requirements.
Trust me, besides your coding/algorithmic skills, you'll also be evaluated
on whether you ask questions to find out requirements for a vaguely defined
problem.
【在 s*******a 的大作中提到】
: but there is always parent node (except for the root) since it is a tree.
: I guess you meant the implementation (e.g., array vs linked data structure
: via pointers), but that should be irrelevant at the algorithmic level..
but in computer science, usually tree is defined as:
struct Node {
Node *left;
Node *right;
};
You cannot assume that each node has a parent node. Actually this is a great
question to ask your interviewer, because you're responsible for finding
out the requirements.
Trust me, besides your coding/algorithmic skills, you'll also be evaluated
on whether you ask questions to find out requirements for a vaguely defined
problem.
【在 s*******a 的大作中提到】
: but there is always parent node (except for the root) since it is a tree.
: I guess you meant the implementation (e.g., array vs linked data structure
: via pointers), but that should be irrelevant at the algorithmic level..
s*a
26 楼
谢谢,明白了。我一般都是习惯从graph的角度想算法,很少从data structure的level
想问题。在我的头脑里algorithm和data structure/code是完全独立的两个东西。可能
是学生的通病吧 :p
这题要是考试的话我会写两个function:一是bottom-up trace-back来找lowest
common ancester, 二是给定一个Node,找他的直接的parent node :p
great
【在 i**********e 的大作中提到】
: for a REAL tree there's a root.
: but in computer science, usually tree is defined as:
: struct Node {
: Node *left;
: Node *right;
: };
: You cannot assume that each node has a parent node. Actually this is a great
: question to ask your interviewer, because you're responsible for finding
: out the requirements.
: Trust me, besides your coding/algorithmic skills, you'll also be evaluated
想问题。在我的头脑里algorithm和data structure/code是完全独立的两个东西。可能
是学生的通病吧 :p
这题要是考试的话我会写两个function:一是bottom-up trace-back来找lowest
common ancester, 二是给定一个Node,找他的直接的parent node :p
great
【在 i**********e 的大作中提到】
: for a REAL tree there's a root.
: but in computer science, usually tree is defined as:
: struct Node {
: Node *left;
: Node *right;
: };
: You cannot assume that each node has a parent node. Actually this is a great
: question to ask your interviewer, because you're responsible for finding
: out the requirements.
: Trust me, besides your coding/algorithmic skills, you'll also be evaluated
H*r
27 楼
Unless the LCA for nodes that's processed has memory otherwise it's still
gona be processed it seems...
// code from the article: ///
Node *LCA(Node *root, Node *p, Node *q) {
if (!root) return NULL;
if (root == p || root == q) return root;
Node *L = LCA(root->left, p, q);
Node *R = LCA(root->right, p, q);
if (L && R) return root; // if p and q are on both sides
return L ? L : R; // either one of p,q is on one side OR p,q is not in L&
R subtrees
}
【在 i**********e 的大作中提到】
: 每个 node 最多只遍历一次。
gona be processed it seems...
// code from the article: ///
Node *LCA(Node *root, Node *p, Node *q) {
if (!root) return NULL;
if (root == p || root == q) return root;
Node *L = LCA(root->left, p, q);
Node *R = LCA(root->right, p, q);
if (L && R) return root; // if p and q are on both sides
return L ? L : R; // either one of p,q is on one side OR p,q is not in L&
R subtrees
}
【在 i**********e 的大作中提到】
: 每个 node 最多只遍历一次。
v*c
28 楼
T(n) = T(m1) + T(m2) + O(1), where m1+m2+1=n
so T(n) = O(n)
L&
【在 H****r 的大作中提到】
: Unless the LCA for nodes that's processed has memory otherwise it's still
: gona be processed it seems...
: // code from the article: ///
: Node *LCA(Node *root, Node *p, Node *q) {
: if (!root) return NULL;
: if (root == p || root == q) return root;
: Node *L = LCA(root->left, p, q);
: Node *R = LCA(root->right, p, q);
: if (L && R) return root; // if p and q are on both sides
: return L ? L : R; // either one of p,q is on one side OR p,q is not in L&
so T(n) = O(n)
L&
【在 H****r 的大作中提到】
: Unless the LCA for nodes that's processed has memory otherwise it's still
: gona be processed it seems...
: // code from the article: ///
: Node *LCA(Node *root, Node *p, Node *q) {
: if (!root) return NULL;
: if (root == p || root == q) return root;
: Node *L = LCA(root->left, p, q);
: Node *R = LCA(root->right, p, q);
: if (L && R) return root; // if p and q are on both sides
: return L ? L : R; // either one of p,q is on one side OR p,q is not in L&
H*r
29 楼
建议加到onlinejudge!
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
【在 i**********e 的大作中提到】
: 面试不会让你写那么复杂算法的。
: 递归程序就只用五行,O(N) 复杂度。
: http://www.leetcode.com/2011/07/lowest-common-ancestor-of-a-bin
a*o
31 楼
怎么找leetcode上面所有的文章啊?
great
【在 i**********e 的大作中提到】
: for a REAL tree there's a root.
: but in computer science, usually tree is defined as:
: struct Node {
: Node *left;
: Node *right;
: };
: You cannot assume that each node has a parent node. Actually this is a great
: question to ask your interviewer, because you're responsible for finding
: out the requirements.
: Trust me, besides your coding/algorithmic skills, you'll also be evaluated
great
【在 i**********e 的大作中提到】
: for a REAL tree there's a root.
: but in computer science, usually tree is defined as:
: struct Node {
: Node *left;
: Node *right;
: };
: You cannot assume that each node has a parent node. Actually this is a great
: question to ask your interviewer, because you're responsible for finding
: out the requirements.
: Trust me, besides your coding/algorithmic skills, you'll also be evaluated
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