以下是使用线段树实现的代码:
#include <stdio.h>
#include <math.h>
#include <limits.h>
#include <iostream>
using namespace std;
int Mid(int s, int e) { return s + (e -s)/2; }
int RMQUtil(int *st, int ss, int se, int qs, int qe, int index) {
if (qs <= ss && qe >= se)
return st[index];
if (se < qs || ss > qe)
return INT_MAX;
int mid = Mid(ss, se);
return min(RMQUtil(st, ss, mid, qs, qe, 2*index+1),
RMQUtil(st, mid+1, se, qs, qe, 2*index+2));
}
int RMQ(int *st, int n, int qs, int qe) {
if (qs < 0 || qe > n-1 || qs > qe)
{
printf("Invalid Input");
return -1;
}
return RMQUtil(st, 0, n-1, qs, qe, 0);
}
int constructSTUtil(int arr[], int ss, int se, int *st, int si) {
if (ss == se) {
st[si] = arr[ss];
return arr[ss];
}
int mid = Mid(ss, se);
st[si] = min(constructSTUtil(arr, ss, mid, st, si*2+1),
constructSTUtil(arr, mid+1, se, st, si*2+2));
return st[si];
}
int *constructST(int arr[], int n) {
int x = (int)(ceil(log2(n)));
int max_size = 2*(int)pow(2, x) - 1;
int *st = new int[max_size];
constructSTUtil(arr, 0, n-1, st, 0);
return st;
}
int main()
{
int arr[] = {1, 2, 3};
int n = sizeof(arr)/sizeof(arr[0]);
int *st = constructST(arr, n);
int qs = 0;
int qe = 2;
int s = 0;
for(int i = 0; i < n; ++i) {
for(int j = 0; j < n - s; ++j) {
cout << RMQ(st, n, j, j + s) << " ";
}
s += 1;
}
cout << endl;
return 0;
}
当然,您可以使用双端队列。找到一种方法,使得最小的元素始终出现在队列的前面,并且队列的大小永远不会超过L。时间复杂度:O(n)