作用：通常求最大(小)子数组/子序列/值

package com.xch.niuke;import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;/*** 滑动窗口** @author XuChenghe* @date 2023/8/27 12:18*/
public class Main001 {public static void main(String[] args) {// 最小值暴力解法int[] res1 = f1(new int[]{3, 1, 4, 5, 2, 6, 7}, 3);System.out.println(Arrays.toString(res1));// 最大值暴力解法int[] res2 = f2(new int[]{3, 1, 4, 5, 2, 6, 7}, 3);System.out.println(Arrays.toString(res2));// 最小值单调队列(递增)解法int[] res3 = f3(new int[]{3, 1, 4, 5, 2, 6, 7}, 3);System.out.println(Arrays.toString(res3));// 最大值单调队列(递减)解法int[] res4 = f4(new int[]{3, 1, 4, 5, 2, 6, 7}, 3);System.out.println(Arrays.toString(res4));}/*** 最小值暴力解法** @param nums* @param k* @return*/public static int[] f1(int[] nums, int k) {int[] res = new int[nums.length - k + 1];for (int i = k - 1; i < nums.length; i++) {res[i - k + 1] = nums[i];for (int j = i; j > i - k; j--) {if (nums[j] < res[i - k + 1]) {res[i - k + 1] = nums[j];}}}return res;}/*** 最大值暴力解法** @param nums* @param k* @return*/public static int[] f2(int[] nums, int k) {int[] res = new int[nums.length - k + 1];for (int i = k - 1; i < nums.length; i++) {res[i - k + 1] = nums[i];for (int j = i; j > i - k; j--) {if (nums[j] > res[i - k + 1]) {res[i - k + 1] = nums[j];}}}return res;}/*** 最小值单调队列(递增)解法** @param nums* @param k* @return*/public static int[] f3(int[] nums, int k) {List<Integer> list = new LinkedList<>();// 用List集合模拟Queue队列int[] res = new int[nums.length - k + 1];for (int i = 0; i < nums.length; i++) {if (i < k - 1) {putAndPop3(null, list, nums[i]);} else if (i == k - 1) {putAndPop3(null, list, nums[i]);res[i - k + 1] = list.get(0);} else {putAndPop3(nums[i - k], list, nums[i]);res[i - k + 1] = list.get(0);}}return res;}/*** 维护单调队列(递增)** @param head* @param list* @param num*/public static void putAndPop3(Integer head, List<Integer> list, int num) {if (head != null && list.get(0) == head) {list.remove(0);}for (int i = list.size() - 1; i >= 0; i--) {if (list.get(i) > num) {// 注意相同的不要清除，否则有异常：-3, 4, 5, 8, 0, -1, 0, 1, 2, 0, 3, 4list.remove(i);} else {break;}}list.add(num);}/*** 最大值单调队列(递减)解法** @param nums* @param k* @return*/public static int[] f4(int[] nums, int k) {List<Integer> list = new LinkedList<>();// 用List集合模拟Queue队列int[] res = new int[nums.length - k + 1];for (int i = 0; i < nums.length; i++) {if (i < k - 1) {putAndPop4(null, list, nums[i]);} else if (i == k - 1) {putAndPop4(null, list, nums[i]);res[i - k + 1] = list.get(0);} else {putAndPop4(nums[i - k], list, nums[i]);res[i - k + 1] = list.get(0);}}return res;}/*** 维护单调队列(递减)** @param head* @param list* @param num*/public static void putAndPop4(Integer head, List<Integer> list, int num) {if (head != null && list.get(0) == head) {list.remove(0);}for (int i = list.size() - 1; i >= 0; i--) {if (list.get(i) < num) {// 注意相同的不要清除，否则有异常list.remove(i);} else {break;}}list.add(num);}}