Description
An array of size n ≤ 106 is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.
|
Window position |
Minimum value |
Maximum value |
|
[1 3 -1] -3 5 3 6 7 |
-1 |
3 |
|
1 [3 -1 -3] 5 3 6 7 |
-3 |
3 |
|
1 3 [-1 -3 5] 3 6 7 |
-3 |
5 |
|
1 3 -1 [-3 5 3] 6 7 |
-3 |
5 |
|
1 3 -1 -3 [5 3 6] 7 |
3 |
6 |
|
1 3 -1 -3 5 [3 6 7] |
3 |
7 |
Your task is to determine the maximum and minimum values in the sliding window at each position.
Input
The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line.
Output
There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.
Sample Input
8 3
1 3 -1 -3 5 3 6 7
Sample Output
-1 -3 -3 -3 3 3
3 3 5 5 6 7
方法一：优先队列
使用STL中的优先队列priority_queue进行求解。开两个优先队列（求max与min），优先队列中插入的是元素的下标，队列中按元素值的从大到小和从小到大优先放在上面。每插入一个元素后，将队列中已经跑出窗口外的元素删除，队列的top元素所指的元素就是该窗口内的最大或最小值。
参考http://hi.baidu.com/xiaohanhoho/blog/item/b020b923aac13cfad6cae228.html
/*优先队列(STL)*/ #include <iostream> #include <vector> #include <queue> #define MAXN 1000005 using namespace std; int S[MAXN];//存数组数据 int Min[MAXN],Max[MAXN];//存最小值和最大值 int cnt; struct cmp1//最小优先队列，结构体+重载运算符，注意掌握写法 { bool operator()(int s1,int s2) { return S[s1]>S[s2]; //元素升序 } }; struct cmp2//最大优先队列 { bool operator()(int s1,int s2) { return S[s1]<S[s2];//元素降序 } }; priority_queue<int,vector<int>,cmp1>Qmin;//自定义优先队列 priority_queue<int,vector<int>,cmp2>Qmax; int main() { int N,K,i; while(scanf("%d%d",&N,&K)!=EOF) { for(i=1; i<=N; i++) scanf("%d",&S[i]); for(i=1; i<=K; i++) //先把第一次查找的前K个插入队列 { Qmin.push(i); Qmax.push(i); } cnt=1; Min[cnt]=S[Qmin.top()];//注意，队列内存的是原始数组的下标 Max[cnt]=S[Qmax.top()]; cnt++; for(i=K+1; i<=N; i++,cnt++) { Qmin.push(i); Qmax.push(i); while(i-Qmin.top()>=K)//删除已经移除届的数 注意此处理解：Qmin.top()为元素最小值的下标，若i与最小值下标间相差大于等于K，即最小值元素已删除，此时应Qmin.pop(); Qmin.pop(); Min[cnt]=S[Qmin.top()]; while(i-Qmax.top()>=K)//删除已经移除届的数 Qmax.pop(); Max[cnt]=S[Qmax.top()]; } for(i=1; i<cnt; i++) printf("%d ",Min[i]); printf("\n"); for(i=1; i<cnt; i++) printf("%d ",Max[i]); printf("\n"); } return 0; }