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Index Sort

/************************************************************************/ /* given an array of integer, write a method to find indices m and n such that if you sorted elements m through n, the entire array would be sorted. Minimize n-m that is find the smallest such sequence eg. input: 1,2,4,7,10,11,12,7,12,6,7,16,18,19 output: (3,10)*/ /* basic idea: using two stacks to remember the numbers, one is from head and the other is from the tail. */ /************************************************************************/ #include <iostream> #include <stack> using namespace std; int getindiceFhead( int a[], int length); int getindiceFtail( int a[], int length); void getIndices( int a[], int length) {        int m, n;        m = getindiceFhead(a, length);        n = getindiceFtail(a, length) - m;        cout<< "(" <<m<< ", " <...
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linked list

3 type of linked list Three different kinds of linked lists exist: Singly linked lists have a single pointer pointing to the next element in the list. The last pointer is empty or points to null, signaling the end of the list. Doubly linked lists have two pointers, one pointing to the next element and one pointing to the previous element. The head node's previous pointer points to null and the tail node's next pointer points to null to signal the end of the list. Circular linked lists usually represent buffers. They have no head or tail, and the primary issue is avoiding infinite traversals because of the cycle. These questions rarely come up in interview questions. Arrays are oftentimes a better substitute for a circular linked list, using the modulus operator to wrap around. 小技巧： 1. 固定head 指针 2. 每次加新节点的时候加在head
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Hanoi Tower

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最近学习到了一个很有趣的problem. 汉诺塔。 我认真的想了好几天，终于在今天找IT给我set up machine等着没事干的时候大概想通了。哈哈哈。 解决这个问题的方法让我想到了binary tree traverse 里面的in order的遍历，汉诺塔的solution之一的 recursive 的方式跟二叉树in order就是异曲同工，而且in human mind, 汉诺塔should make more sense to understand compared to binary tree. 在网上看到了 一个印度小男孩对汉诺塔的分析 ，果然是tree. Basically 解决汉诺塔的问题就是一个搬家公司应该怎么搬家的问题，from mission completed to start, which is from bottom to top. Let's define Hanoi move function as: HanoiMove(int leftToSit, int from - 0 , int to - 1, int rest - 2) and we have n pieces in total waiting for the move In theory, to approach the last step nth of move, we just need finish move rest of Hanoi pieces (n-1 in total) from start to rest pile,(简单说来就是，清空路障，直捣虎穴。）and then it should be able to move the last piece from start to To pile. so we just do 1) HanoiMove(n-1,  from, res, to)    ---->>>    left as we planned in theory, we move the nth piece move(from,  to)    after this move, we have...
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