用两个栈实现队列&&旋转非递减数列的最小值

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用两个栈实现队列，比较基础的一道题目了，关键是template的运用。代码如下

#include<iostream>#include<string>#include<stack>using namespace std;template<typename T>class MyQueue{public:void push(T n){a.push(n);}void pop(){if(!b.empty())b.pop();else if(!a.empty()){while(!a.empty()){T n = a.top();a.pop();b.push(n);}b.pop();}throw new exception("queue is empty");}T head(){if(!b.empty())return b.top();else if(!a.empty()){while(!a.empty()){T n = a.top();a.pop();b.push(n);}return b.top();}throw new exception("queue is empty");}bool empty(){if(a.empty()&&b.empty())return true;return false;}private:stack<T> a;stack<T> b;};int main(){MyQueue<int> myqueue;myqueue.push(1);myqueue.push(2);myqueue.push(3);myqueue.push(4);while(!myqueue.empty()){int n = myqueue.head();printf("%d",n);myqueue.pop();}myqueue.pop();system("PAUSE");return 0;}

旋转非递减数列的最小值。既然是非递减数列，可以考虑二分搜索，旋转之后有两个非递减序列，最小值在后面的递减序列里，一步步的缩小范围。但要注意二分搜索应用时递增序列，如果beg,mid,end三个位置对应的数组值相同，则没法划分范围，必须顺序查找。

这样就是log(n)的解法，当然最简单的o(n)也很容易，就是按顺序查找。代码如下

#include<iostream>#include<string>#include<stack>using namespace std;int getorder(int a[],int beg,int end){int mina = a[beg];for(int i = beg+1;i<=end;i++)if(a[i]<mina)mina = a[i];return mina;}int getmin(int a[],int length){int beg = 0;int end = length-1;int mid = beg;while(a[beg] >= a[end]){if(end-beg == 1){mid = end;break;}mid = (beg+end)>>1;if(a[beg] == a[mid] && a[mid] == a[end]){return getorder(a,beg,end);}if(a[mid] >= a[beg])beg = mid;else if(a[mid] <= a[end])end = mid;}return a[mid];}int main(){int a[] = {4,5,6,7,8,1,2,3};int n = getmin(a,8);printf("%d\n",n);system("PAUSE");return 0;}

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