亚洲第一二三四区,久久久久亚洲精品无码网址,久久久久亚洲AV无码专区桃色

我是如何�q�入��h��

Meng Lee — Thu, 12 Jun 2014 18:00:00 GMT

今天�Q�终于有旉��静下心来回顾�q�去两年来所做的事情�Q�感慨万千，一时之间竟不知从何说�v。两�q�以来，遇到的困隄��实不��，但每每遭遇挫折与不顺之后�Q�却往往能柳暗花明，遇到新的转机�Q?/span>让我真真切切地感受到了功夫不负有心�h�q�句话的含意�?/span>

一、�ؓ什么要出国
其实�Q�之前从来没有考虑�q�要出国�Q�更没有惌��能直接出国工作�?/span>回想一下，�q�个军_��的做出，多半�q�是�~�于自己骨子里的不安分�?/span>我从很大�E�度上来说是一个闲不住的�h�Q�从��学、中学�?/span>大学到研�I�生�Q�我几乎每天都有明确的目标。然而，2013�q�从公司��C��业单位工作以后，我的生活发生了巨大地转变。简单的工作、空�z�的公文�?/span>无聊的活动占据了我全部的工作��d��?/span>有段旉��几乎天天写材料搞�z�d��。领导经常夸我材料写得又快又好，�z�d��也搞得有声有�Ԍ��心里感觉很有成就感。然而，旉��一长，逐渐发现那些公文永远是一个套路，以至于我分门别类�Q?/span>摸烦��Z��几个万能模板。而活动则千篇一律，让�h疲于应付�?/span>我甚臛_��以看到六十岁退休时我在�q�什么，于是一阉|��惧感常常会莫名袭来，因�ؓ我不安分、不满��于此�?/span>我不能放弃所学所长，我不能庸庸碌��在�q�里度过未来的几十年�Q?/span>我还有梦惻I��我还要登高看世界。�ؓ了这个，我走�q�了不��^凡的两年�?/span>

二、如何出�?/span>
对于普通�h来说�Q�出国大致有三条路�?/span>
�W�一条�\是申请去国外留学�Q?/span>取得学位之后以应届毕业生的��n份找工作�Q�然后留在国外生�z�R�?/span>�q�是一条比较稳妥、简便的路，走这条�\的�h最多�?/span>
�W�二条�\是先�q�入跨国公司的中国分公司工作一�D�|��_��然后找机会外�z�ֈ�国外总部工作。走�q�条路的要求比较多，首先要能够进入比较大的跨国公司工作，其次�q�个公司愿意��中国员工transfer到国外，同时�q�要外国总部有部门愿意接收你�Q�所以还是需要一些运气�?/span>但是�Q�如果成功，好处也显而易见。省��M��M��的时间和学费�Q?/span>降低了家庭负担，对于家境一般的人是非常好的选择�?/span>
�W�三条�\是直接参加外国公司的面试�Q�通过之后直接��d��外工作�?/span>�q�条路要求最高，需要通过外国公司严格的面试，另外�q�要能够成功取得�{�证�Q�美国工作签证就需要抽�{�）。因此，走这条�\更需要实力、机遇和�q�气�?/span>
鉴于�W�三条�\非常难走�Q��ؓ了保证成功，我选择了同时申请学校和参加外国公司面试的办法，�q�也注定了我��付出更多的艰苦努力�?/span>

三、申请学�?/span>
甌��学校从准备到最�l�完成，我大概用了一�q�时间�?/span>光��参加了三�ơGRE和一�ơ托��考试。回惛_��备的�q�程�Q?/span>最大的敌�h��是自己�Q�最重要的法宝就是坚持坚持再坚持�?/span>记得�W�一�ơ考GRE没有取得理想的成�l�，因�ؓ是第一�ơ参加英语考试�Q�心情非常失落。幸亏当时有��x��友（现在的老婆�Q�的鼓励�Q�我�l�箋复习没有攑ּ�。经�q�一个月的复习，取得了非�怸�错的托福成�W。记得托��出成�W的那天，我紧张得不敢查，点开��面的那一刻，我都不敢�怿�居然能有�q�么不错的成�l�。特别是听力�Q?/span>考试的时候觉得好几个都没有听清楚�Q�最后居然有27分，真是不可思议�Q�可见功夫不负有心�h�Q�付出��L��回报的�?/span>
有了��p��成�W之后�Q�就是撰写申��h��书。这斚w��我完全没有经验，所有的信息全部是通过一亩三分地论坛获得的�?/span>�q�个论坛信息非常丰富�Q�基本上所有申��L��关的内容都有涉及�?/span>我每天都会花一些时间浏览别人的帖子�Q��ؓ自己定位选校�Q?/span>找文书灵感等�{�。非常感谢我的本�U�和研究生导师，�q�有蒋志诚�ؓ我递推荐信�Q�没有你们的帮助�Q?/span>我不可能完成甌��工作�?/span>
最后，我申请了��国和加拿大的十五所学校的计��机专业的研�I�生�Q?/span>拿到了CMU、USC和多伦多大学的offer。其中，CMU的Data Science program应该是世界数一��C��的，录取率非�怽��Q?/span>毕业后的��d��也非常好�Q�大多数都可以进入美国一��公司工作�?/span>多大也是加拿大排名第一的学校，计算机的��׃��也非常好�?/span>

四、Facebook的面�?/span>
参加Facebook的面试也完全是无意的�Q?/span>在Linkedin上收��C��Facebook HR的邀请信�Q�于是也没有怎么准备��做了电面，居然反馈非常好，马上��q��我安排了onsite面试�Q�地�Ҏ��印度的�v得拉巴�?/span>但是�Q�始�l�是没有做什么准备，而且和谷歌不一��L��是，HR办事效率实在太高�Q�每一轮间隔都非常短，��D��我根本没有时间热�w�一下，�q�leetcode都没有做�q�就匆匆参加面试了，最�l�没有如愉K��过面试�?/span>
不过�Q�这�ơ面试还是很有收莗��第一�ơ出国，�W�一�ơ参加美国公司全英文面试�Q�学��C��太多�Q�积累了�l�验�Q?/span>可以说如果没有Facebook的失败，我是不可能进入谷歌的�?/span>

五、Google的面�?/span>
参加��h��的面试可以说完全是老婆的怂恿�?/span>从印度参加完Facebook面试回来之后�Q?/span>我就开始专心于学校甌��了。但是，老婆��我试试面一下Google�?/span>�׃��Facebook的失利和Google�q�乎苛刻的面试流�E�，我开始是抗拒参加的。最后，在老婆的一再要求下�Q?/span>我终于找了一个在��h��上�v工作的师兄做了内推�?/span>四月底我收到了谷歌北京HR的第一通电话，也正式拉开了我为期一�q�的面试��程�?/span>
和HR通电话不久，我进行了�W�一�ơ电话面试�?/span>��h��的电话面试和Facebook差不多，��是面试官打�q�来�Q?/span>把题目口�q��ƈ且写在Google Doc上，然后我把�E�序写在Google Doc上。第一�ơ电面的题目不难�Q?/span>但谷歌对代码效率和清晰度的要求远�q�超��Z��我的惛_��?/span>�W�一轮面得磕��绊�l�，但是�q�好面试官是中国人，非常nice�Q?/span>没有让我fail�?/span>
于是�Q�我又被要求�q�行�W�二�ơ电面。期间由于面试官临时有事爽约�Q?/span>我等了差不多一个月。但是，也就是这一个月�Q?/span>我努力做了一些准备，虽然面试依旧不是十全十美�Q?/span>但是我还是有惊无险地�q�入��C��onsite面试环节�?/span>
虽然可以onsite面试了，但是我依旧对�q�入��h��不报��M��希望�Q�因为我清楚的知道，��h��面试实在是太难了�Q�onsite面试的挑战将�q�远大于电面�?/span>因此�Q�我��d��京面试完全是惛_��一�ơ免�Ҏ��游�?/span>面试前一天还�怹�不见的万威夫妇吃饭，聊得很开心，完全没有把面试放在心上�?/span>
也许是放杄��原因�Q�我前一天晚上睡得很好，�W�二天我�_��非常好�?/span>
不过��h��毕竟是谷歌，面试�W�一轮一上来��q��了我一个下马威�?/span>一个coding题一个设计题�Q�表面上很简单，但是做出来��L��有这样那��L��问题�Q?/span>�W�一轮完了之后我基本打算回家了�?/span>
但是�Q�不知道��Z��么，从第二轮开始，��p��来越��利�Q?/span>coding做得非常好，基本上是一�ơ写完，没有涂改�Q?/span>也没有被面试官找到大的bug。突然之��_��隐隐感觉出现了一丝希望�?/span>
四轮�q�后�Q�我�l�束了第一�ơonsite面试。但是，三天之后�Q?/span>我被告知�׃��设计题做得不好，我被要求�q�行一�ơ加面，地点在上��于是，我又在上��做了一�ơ面试，只有一个设计题�?/span>我感觉答得还可以�Q�但是心情真的是忐忑不安�Q?/span>特别是接下来的一个礼拜，几乎是坐立不安�?/span>
记得是一个礼拜之后的�C�拜五中午，我正做准备主持下午的道�d讲堂�Q�突然接��C��一�?10的电话，我知道是��h��的电话。接通电话的那一刻，�I�气都几乎要凝固了，当听到通过HC的消息时�Q�我�Ȁ动得不能自已�?/span>不可能完成的��d��居然完成了，虽然不知道能不能�ȝ��国总部工作�Q?/span>但是能进入谷歌已�l�非�怸��Ҏ��了，而且��h��非常鼓励transfer�ȝ��国工作，因此��Z��q�是很多的�?/span>
然而，让我没有惛_��的是�Q�接下来的team match却异常艰难，陆陆�l�箋几个team都没有成功match上�?/span>转眼��到�?014�q�春季，半年的等待让我对何时�q�入��h��非常悲观�Q?/span>加上甌��学校工作十分�J�重�Q�我基本没有��x��q�个事情�?/span>
��在我快要放弃的时候，拿到了美国一个公司的offer�Q?/span>他们�{�应�l�我办H1B�{�证。于是，我把�q�个情况告诉了谷歌，要求他们��快�l�找到team�Q�不然我��去��国了�?/span>�l�果��h��居然在三天之内�ؓ我match上了英国office的一个team�Q�让��Z��得不感叹�q�是要offer多才好啊�Q�于是，我又�l�过了近三个月的�{�证办理��程�Q�终于要启程赴英了�?/span>

回顾两年来的努力�Q�终于要实现自己的梦想了�Q�感慨万千�?/span>在短短的人生中，能有�q�一�D�不��d��的经历，我觉得十分幸�q��?/span>展望未来�Q�我惌��万卷书不如行万里路，未来希望能够利用在��u敦工作的��Z��Q�尽量多��L��z�各国走赎ͼ�丰富自己的阅历，开拓自��q��眼界�?/span>

最后要感谢老婆一直以来的支持和鼓励，你一直是我前�q�的动力�Q?/span>其次要感谢父母的不理解和不支持，你们的反对让我更加完善了自己的计划，逼着我找��C��一条最好的��\�Q�还要感谢师长和朋友们的帮助�Q?/span>感谢杨老师和沈老师�q�有蒋志诚不厌其烦地帮我递推荐信�Q?/span>感谢万威夫妇多次在北京款待我�Q�没有你们的��食�Q?/span>我是不可能完成面试的�Q�还有许多帮助过我的人，在这里就不能一一感谢了�?/span>

Meng Lee 2014-06-13 02:00 发表评论

[Leetcode] Trapping Rain Water

Meng Lee — Tue, 14 Jan 2014 01:16:00 GMT

��法很简单，核心思想是：�Ҏ��个值A[i]来说�Q�能trapped的最多的water取决于在i之前最高的值leftMostHeight[i]和在i双��的最高的值rightMostHeight[i]。（均不包含自��n�Q�。如果min(left,right) > A[i]�Q�那么在i�q�个位置上能trapped的water��是min(left,right) – A[i]�?/div>

有了�q�个��x��好办了�Q�第一遍从左到双��数�l�leftMostHeight�Q�第二遍从右到左计算rightMostHeight�Q�在�W�二遍的同时��可以计��出i位置的结果了�Q�而且�q�不需要开�I�间来存放rightMostHeight数组�?/div>

旉��复杂度是O(n)�Q�只扫了两遍�?br />

1 public class TrappingRainWater {
2     public int trap(int A[], int n) {
3         if (n <= 2)
4             return 0;
5
6         int[] lmh = new int[n];
7         lmh[0] = 0;
8         int maxh = A[0];
9         for (int i = 1; i < n; ++i) {
10             lmh[i] = maxh;
11             if (maxh < A[i])
12                 maxh = A[i];
13         }
14         int trapped = 0;
15         maxh = A[n - 1];
16         for (int i = n - 2; i > 0; --i) {
17             int left = lmh[i];
18             int right = maxh;
19             int container = Math.min(left, right);
20             if (container > A[i]) {
21                 trapped += container - A[i];
22             }
23             if (maxh < A[i])
24                 maxh = A[i];
25         }
26         return trapped;
27     }
28 }

Meng Lee 2014-01-14 09:16 发表评论

[Leetcode] Permutation Sequence

Meng Lee — Tue, 07 Jan 2014 08:06:00 GMT

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,

We get the following sequence (ie, for n = 3):

"123"

"132"

"213"

"231"

"312"

"321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

�q�道题其实有很强的规律可循。首先，n个元素的排列��L��是n!。在下面的分析中�Q�让k的范围是0 <= k < n!。（题目代码实际上是1<=k<=n!)
可以看到一个规律，��是�q�n�Q�个排列中，�W�一位的元素��L��(n-1)!一�l�出现的�Q�也��p��如果p = k / (n-1)!�Q�那么排列的最开始一个元素一定是arr[p]�?/div>

�q�个规律可以�c�L��下去�Q�在剩余的n-1个元素中逐渐挑选出�W�二个，�W�三个，...�Q�到�W�n个元素。程序就�l�束�?/div>

1 /**
2  * The set [1,2,3,…,n] contains a total of n! unique permutations.
3  *
4  * By listing and labeling all of the permutations in order, We get the
5  * following sequence (ie, for n = 3):
6  *
7  * "123" "132" "213" "231" "312" "321" Given n and k, return the kth permutation
8  * sequence.
9  *
10  * Note: Given n will be between 1 and 9 inclusive.
11  *
12  */
13
14 public class PermutationSequence {
15     public String getPermutation(int n, int k) {
16         char[] arr = new char[n];
17         int pro = 1;
18         for (int i = 0; i < n; ++i) {
19             arr[i] = (char) ('1' + i);
20             pro *= (i + 1);
21         }
22         k = k - 1;
23         k %= pro;
24         pro /= n;
25         for (int i = 0; i < n - 1; ++i) {
26             int selectI = k / pro;
27             k = k % pro;
28             pro /= (n - i - 1);
29             int temp = arr[selectI + i];
30             for (int j = selectI; j > 0; --j) {
31                 arr[i + j] = arr[i + j - 1];
32             }
33             arr[i] = (char) temp;
34         }
35         return String.valueOf(arr);
36     }
37 }
38

Meng Lee 2014-01-07 16:06 发表评论

[Leetcode] Lowest Common Ancestor of a Binary Search Tree (BST) && Lowest Common Ancestor of Binary Tree

Meng Lee — Mon, 06 Jan 2014 01:32:00 GMT

Lowest Common Ancestor of a Binary Search Tree (BST)

Given a binary search tree (BST), find the lowest common ancestor of two given nodes in the BST.

_______6______

/ \

___2__ ___8__

/ \ / \

0 _4 7 9

/ \

3 5

Using the above tree as an example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6.

But how about LCA of nodes 2 and 4? Should it be 6 or 2?

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between

two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a

node to be a descendant of itself).” Since a node can be a descendant of itself, the LCA of 2 and

4 should be 2, according to this definition.

Hint:

A top-down walk from the root of the tree is enough.

1 public class LowestCommonAncestorOfaBinarySearchTree {
2     public TreeNode LCA(TreeNode root, TreeNode p, TreeNode q) {
3         if (root == null || p == null || q == null)
4             return null;
5         if (Math.max(p.val, q.val) < root.val)
6             return LCA(root.left, p, q);
7         if (Math.min(p.val, q.val) > root.val)
8             return LCA(root.right, p, q);
9         return root;
10     }
11 }

Given a binary tree, find the lowest common ancestor of two given nodes in the tree.

_______3______

/ \

___5__ ___1__

/ \ / \

6 _2 0 8

/ \

7 4

If you are not so sure about the definition of lowest common ancestor (LCA), please refer to my previous

post: Lowest Common Ancestor of a Binary Search Tree (BST) or the definition of LCA here. Using the tree

above as an example, the LCA of nodes 5 and 1 is 3. Please note that LCA for nodes 5 and 4 is 5.

Hint:

Top-down or bottom-up? Consider both approaches and see which one is more efficient.

1 public class LowestCommonAncestorOfBinaryTree {
2     public TreeNode LCA(TreeNode root, TreeNode p, TreeNode q) {
3         if (root == null)
4             return null;
5         if (root == p || root == q)
6             return root;
7         TreeNode left = LCA(root.left, p, q);
8         TreeNode right = LCA(root.right, p, q);
9         if (left != null && right != null)
10             return root;
11         return left != null ? left : right;
12     }
13 }

Meng Lee 2014-01-06 09:32 发表评论

[Leetcode] Scramble String

Meng Lee — Sun, 05 Jan 2014 04:33:00 GMT

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

great

/ \

gr eat

/ \ / \

g r e at

/ \

a t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

rgeat

/ \

rg eat

/ \ / \

r g e at

/ \

a t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

rgtae

/ \

rg tae

/ \ / \

r g ta e

/ \

t a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

1 public class ScrambleString {
2     public boolean isScramble(String s1, String s2) {
3         if (s1.length() != s2.length())
4             return false;
5         if (s1.equals(s2))
6             return true;
7
8         int[] A = new int[26];
9         for (int i = 0; i < s1.length(); i++) {
10             ++A[s1.charAt(i) - 'a'];
11         }
12
13         for (int j = 0; j < s2.length(); j++) {
14             --A[s2.charAt(j) - 'a'];
15         }
16
17         for (int k = 0; k < 26; k++) {
18             if (A[k] != 0)
19                 return false;
20         }
21
22         for (int i = 1; i < s1.length(); i++) {
23             boolean result = isScramble(s1.substring(0, i), s2.substring(0, i))
24                     && isScramble(s1.substring(i), s2.substring(i));
25             result = result
26                     || (isScramble(s1.substring(0, i),
27                             s2.substring(s2.length() - i, s2.length())) && isScramble(
28                             s1.substring(i), s2.substring(0, s2.length() - i)));
29             if (result)
30                 return true;
31         }
32         return false;
33     }
34 }

Meng Lee 2014-01-05 12:33 发表评论

[Leetcode] Largest Rectangle in Histogram

Meng Lee — Sun, 05 Jan 2014 04:31:00 GMT

Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.

Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].

The largest rectangle is shown in the shaded area, which has area = 10 unit.

For example,

Given height = [2,1,5,6,2,3],

return 10.

本题需要��用栈�l�护一个高度单调递增的序列下标，如果遇到一个元素比当前栈顶元素高度��，那么出栈�Q��ƈ计算当前最大面�U�。如果栈为空�Q�需要特�D�考虑�?br />

1 public class LargestRectangleinHistogram {
2     public int largestRectangleArea(int[] height) {
3         Stack stack = new Stack();
4         int i = 0;
5         int maxArea = 0;
6         int[] h = new int[height.length + 1];
7         h = Arrays.copyOf(height, height.length + 1);
8         while (i < h.length) {
9             if (stack.isEmpty() || h[stack.peek()] <= h[i]) {
10                 stack.push(i++);
11             } else {
12                 int t = stack.pop();
13                 maxArea = Math.max(maxArea, h[t] * (stack.isEmpty() ? i : i - stack.peek() - 1));
14             }
15         }
16         return maxArea;
17     }
18 }

Meng Lee 2014-01-05 12:31 发表评论

[Leetcode] Binary Tree Inorder Traversal

Meng Lee — Sat, 04 Jan 2014 03:17:00 GMT

Given a binary tree, return the inorder traversal of its nodes' values.

For example:

Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

切记p节点初始时指向root.left。代码如下：

1 public class BinaryTreeInorderTraversal {
2     public ArrayList inorderTraversal(TreeNode root) {
3         ArrayList inOrder = new ArrayList();
4         if (root == null)
5             return inOrder;
6         Stack s = new Stack();
7         s.add(root);
8         TreeNode p = root.left;
9         while (!s.empty()) {
10             while (p != null) {
11                 s.add(p);
12                 p = p.left;
13             }
14             TreeNode n = s.pop();
15             inOrder.add(n.val);
16             p = n.right;
17             if (p != null) {
18                 s.add(p);
19                 p = p.left;
20             }
21         }
22         return inOrder;
23     }
24 }

Meng Lee 2014-01-04 11:17 发表评论

[Leetcode] Subsets II

Meng Lee — Fri, 03 Jan 2014 08:40:00 GMT

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

Elements in a subset must be in non-descending order.

The solution set must not contain duplicate subsets.

For example,

If S = [1,2,2], a solution is:

[

[2],

[1],

[1,2,2],

[2,2],

[1,2],

[]

]

�׃��元素中可能存在重复，因此较之于Subset的实玎ͼ�需要加一些判断。如果碰��C��重复元素�Q�只需要在上一�ơ�P代新增的子集的基��上再�q�行�q�代卛_��。实��C��码如下：

1 public class SubsetsII {
2     public ArrayList> subsetsWithDup(int[] num) {
3         ArrayList> ret = new ArrayList>();
4         ArrayList> lastLevel = null;
5         ret.add(new ArrayList());
6         Arrays.sort(num);
7         for (int i = 0; i < num.length; i++) {
8             ArrayList> tmp = new ArrayList>();
9             ArrayList> prev = i == 0 || num[i] != num[i - 1] ? ret : lastLevel;
10             for (ArrayList s : prev) {
11                 ArrayList newSet = new ArrayList(s);
12                 newSet.add(num[i]);
13                 tmp.add(newSet);
14             }
15             ret.addAll(tmp);
16             lastLevel = tmp;
17         }
18         return ret;
19     }
20 }

Meng Lee 2014-01-03 16:40 发表评论

[Leetcode] Word Ladder II

Meng Lee — Thu, 02 Jan 2014 05:59:00 GMT

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that:

Only one letter can be changed at a time

Each intermediate word must exist in the dictionary

For example,

Given:

start = "hit"

end = "cog"

dict = ["hot","dot","dog","lot","log"]

Return

[

["hit","hot","dot","dog","cog"],

["hit","hot","lot","log","cog"]

]

Note:

All words have the same length.

All words contain only lowercase alphabetic characters.

�q�个题目应该��是leetcode上比较难的题目了。刚开始我采用了和Word Ladder�怼�的做法，只是用ArrayList记录了当前变换�\径，在小数据的情况下可以Accept�Q�但是大数据��时。究其原因，是由于�ؓ每个当前节点记录变换路径的时候，需要复制之前的ArrayList�Q�这个时间开销较大�?br />其实�Q�我们可以采用一个Map>�l�构�Q�记录字典单词的每一个前驱，�q�样我们可以从end反向遍历�Q�构造出转换路径�?br />同时�Q�我利用了两个ArrayList�Q�交替记录上一层和下一层的节点�Q�如果下一层节点�ؓ�I�，则不存在路径�Q�立卌��回。如果下一层中出现了end�Q�证明找��C��所有的最短�\径，停止搜烦开始构造�\径。实��C��码如下：

1 public class WordLadderII {
2     private void GeneratePath(Map> prevMap,
3             ArrayList path, String word,
4             ArrayList> ret) {
5         if (prevMap.get(word).size() == 0) {
6             path.add(0, word);
7             ArrayList curPath = new ArrayList(path);
8             ret.add(curPath);
9             path.remove(0);
10             return;
11         }
12
13         path.add(0, word);
14         for (String pt : prevMap.get(word)) {
15             GeneratePath(prevMap, path, pt, ret);
16         }
17         path.remove(0);
18     }
19
20     public ArrayList> findLadders(String start, String end,
21             HashSet dict) {
22         ArrayList> ret = new ArrayList>();
23         Map> prevMap = new HashMap>();
24         dict.add(start);
25         dict.add(end);
26         for (String d : dict) {
27             prevMap.put(d, new ArrayList());
28         }
29         ArrayList> candidates = new ArrayList>();
30         candidates.add(new HashSet());
31         candidates.add(new HashSet());
32         int current = 0;
33         int previous = 1;
34         candidates.get(current).add(start);
35         while (true) {
36             current = current == 0 ? 1 : 0;
37             previous = previous == 0 ? 1 : 0;
38             for (String d : candidates.get(previous)) {
39                 dict.remove(d);
40             }
41             candidates.get(current).clear();
42             for (String wd : candidates.get(previous)) {
43                 for (int pos = 0; pos < wd.length(); ++pos) {
44                     StringBuffer word = new StringBuffer(wd);
45                     for (int i = 'a'; i <= 'z'; ++i) {
46                         if (wd.charAt(pos) == i) {
47                             continue;
48                         }
49
50                         word.setCharAt(pos, (char) i);
51
52                         if (dict.contains(word.toString())) {
53                             prevMap.get(word.toString()).add(wd);
54                             candidates.get(current).add(word.toString());
55                         }
56                     }
57                 }
58             }
59
60             if (candidates.get(current).size() == 0) {
61                 return ret;
62             }
63             if (candidates.get(current).contains(end)) {
64                 break;
65             }
66         }
67
68         ArrayList path = new ArrayList();
69         GeneratePath(prevMap, path, end, ret);
70
71         return ret;
72     }
73 }

Meng Lee 2014-01-02 13:59 发表评论