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懒甜儿

Monday, August 18, 2014. Leetcode - Permutations I & II. Given a collection of numbers, return all possible permutations. Have the following permutations:. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Have the following unique permutations:. 0 & num[i] = num[i - 1] & visited[i - 1] = 0) { continue; } visited[i] = 1; list.add(num[i]); helper(result, list, visited, num); list.remove(list.size() - 1); visited[i] = 0; } } }. Wednesday, August 13, 2014.

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懒甜儿 | ttianzhao.blogspot.com Reviews
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Monday, August 18, 2014. Leetcode - Permutations I & II. Given a collection of numbers, return all possible permutations. Have the following permutations:. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Have the following unique permutations:. 0 & num[i] = num[i - 1] & visited[i - 1] = 0) { continue; } visited[i] = 1; list.add(num[i]); helper(result, list, visited, num); list.remove(list.size() - 1); visited[i] = 0; } } }. Wednesday, August 13, 2014.
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for example,thoughts,posted by,tia zhao,no comments,email this,blogthis,share to twitter,share to facebook,share to pinterest,支持removerandom的hashtable,labels 实现题,实现hashmap,若是thread safe,用synchronized就ok了,flip tree,题目如下:,labels 面经,两个玩家取数,应该有dp解法,滑动窗口 单调队列
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懒甜儿 | ttianzhao.blogspot.com Reviews

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Monday, August 18, 2014. Leetcode - Permutations I & II. Given a collection of numbers, return all possible permutations. Have the following permutations:. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Have the following unique permutations:. 0 & num[i] = num[i - 1] & visited[i - 1] = 0) { continue; } visited[i] = 1; list.add(num[i]); helper(result, list, visited, num); list.remove(list.size() - 1); visited[i] = 0; } } }. Wednesday, August 13, 2014.

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1

懒甜儿: July 2014

http://ttianzhao.blogspot.com/2014_07_01_archive.html

Wednesday, July 30, 2014. 看好几个面经有这题,有些面经写的不清楚还以为是word ladder呢, 终于有个大大写的特别清楚。 面经转载于:http:/ www.mitbbs.com/clubarticle t/New Mommy and New Daddy/20337627.html. 给一个words list, 输入两个单词,找出这两个单词在list中的最近距离(先. Green', 'blue', 'orange', 'purple', 'green'] f.distance(list, 'blue', '. Green') # output 1. 面经上看来的题,Leetcode上有Clone Graph 和 Copy List with Random Pointer 都是生成一个结构的镜像,但是因为二叉树本身结构的原因,这个词比Leetcode上的两个题都要简单。 Labels: Tree and Graph. Wednesday, July 23, 2014. Concurrent - 实现一个blocking queue. Concurrent - H2O 问题.

2

懒甜儿: 求数组中出现次数超过一半的元素 || 从有n个元素的数组中找出出现次数大于n/k次的元素

http://ttianzhao.blogspot.com/2014/08/blog-post.html

Sunday, August 3, 2014. 给定一个数组,有一个元素出现超过1/2次,找出这个元素。 Public int findElem(int[] arr){ if(arr.length= 0) return Integer.MIN VALUE; int cnt = 1; int elem = arr[0]; for(int i=1; i arr.length; i ){ if(arr[i]= elem) cnt ; else if(cnt 0 & arr[i]! Elem) cnt- ; else{ cnt = 1; elem = arr[i]; } } return cnt 0? Elem : Integer.MIN VALUE; }. 衍生题:给定一个数组,有两个元素出现超过1/3次,找出这两个元素。 以下解题思路我从mitbbs坛子上看到某大大写的,但我复制完就找不到链接了,如果大大你看到请谅解,我不是故意不标明出处的。 2,实现的时候,可以用hash表保存每个数字出现的次数,当hash表中节点个数为3时,. K2 : k3) : k1; k2 = v2= 0?

3

懒甜儿: August 2014

http://ttianzhao.blogspot.com/2014_08_01_archive.html

Monday, August 18, 2014. Leetcode - Permutations I & II. Given a collection of numbers, return all possible permutations. Have the following permutations:. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Have the following unique permutations:. 0 & num[i] = num[i - 1] & visited[i - 1] = 0) { continue; } visited[i] = 1; list.add(num[i]); helper(result, list, visited, num); list.remove(list.size() - 1); visited[i] = 0; } } }. Wednesday, August 13, 2014.

4

懒甜儿: Leetcode - Permutations I && II

http://ttianzhao.blogspot.com/2014/08/leetcode-permutations-i-ii.html

Monday, August 18, 2014. Leetcode - Permutations I & II. Given a collection of numbers, return all possible permutations. Have the following permutations:. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Have the following unique permutations:. 0 & num[i] = num[i - 1] & visited[i - 1] = 0) { continue; } visited[i] = 1; list.add(num[i]); helper(result, list, visited, num); list.remove(list.size() - 1); visited[i] = 0; } } }. PrintFactors & Two Sum.

5

懒甜儿: 实现list interface

http://ttianzhao.blogspot.com/2014/08/list-interface.html

Monday, August 11, 2014. 又是面经啊,不知道意义何在,摘从library:. 对于list这个借口,要注意的是list里的元素是有序的,可以按index存,取. Package java.util; public interface List E extends Collection E { int size(); boolean isEmpty(); / Returns an array containing all of the elements in this list in proper sequence (from first to last element). Object[] toArray(); T T[] toArray(T[] a); / Appends the specified element to the end of this list (optional operation). boolean add(E e); void add(int index, E element); boolean addAll(Collection?

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懒甜儿

Monday, August 18, 2014. Leetcode - Permutations I & II. Given a collection of numbers, return all possible permutations. Have the following permutations:. Given a collection of numbers that might contain duplicates, return all possible unique permutations. Have the following unique permutations:. 0 & num[i] = num[i - 1] & visited[i - 1] = 0) { continue; } visited[i] = 1; list.add(num[i]); helper(result, list, visited, num); list.remove(list.size() - 1); visited[i] = 0; } } }. Wednesday, August 13, 2014.

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