【问题】

给定数组arr和整数 n u m num num,返回共有多少个子数组满足如下情况:
m a x ( a r r [ i . . . j ] ) m i n ( a r r [ i . . . j ] ) &lt; = n u m max(arr[i...j])-min(arr[i...j])&lt;=num max(arr[i...j])min(arr[i...j])<=num

【解答】

package com.chanmufeng.codingInterviewGuide.stackAndQueue_10;

import java.util.LinkedList;

/**
 *
 */
public class SubArrNum {
    public static int getNum(int[] arr, int num) {
        if (arr == null || arr.length == 0)
            return 0;

        int L = 0;
        int R = 0;
        int res = 0;
        LinkedList<Integer> qmax = new LinkedList<>();
        LinkedList<Integer> qmin = new LinkedList<>();

        while (L < arr.length) {
            while (R < arr.length) {
                while (!qmax.isEmpty() && arr[R] >= arr[qmax.peekLast()]) {
                    qmax.pollLast();
                }
                qmax.addLast(R);

                while (!qmin.isEmpty() && arr[R] <= arr[qmin.peekLast()]) {
                    qmin.pollLast();
                }
                qmin.addLast(R);

                if (arr[qmax.peekFirst()] - arr[qmin.peekFirst()] > num) {
                    break;
                }
                R++;
            }
            res += (R - L);

            if (qmax.peekFirst() == L) {
                qmax.pollFirst();
            }
            if (qmin.peekFirst() == L) {
                qmin.pollFirst();
            }
            L++;
        }
        return res;
    }

    public static void main(String[] args) {
        int[] arr = new int[]{1, 2, 3};
        System.out.println(getNum(arr, 1));
    }
}