算法基础提升——滑动窗口、单调栈和单调栈的应用

package com.zuoshen.jichutisheng.class04;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Stack;

/**
 * @author ShiZhe
 * @create 2022-03-28 19:45
 */
public class code01 {

    /**
     * 滑动窗口
     * @param arr
     * @param w 窗口大小
     * @return
     */
    public static int[] slidingWindow(int[] arr, int w) {
        if (arr == null || arr.length < w || w < 1) {
            return null;
        }
        // 存放的是下标，头是最大值，尾是最小值。双端队列
        LinkedList<Integer> indexList = new LinkedList<>();
        // 窗口最大值的数组
        int[] res = new int[arr.length - w + 1];
        int index = 0;
        for (int i = 0; i < arr.length; i++) {
            // 当队列不为空时，peekLast拿到队列最小值（尾部）的下标,与当前i的值比较，小于弹出
            while (!indexList.isEmpty() && arr[indexList.peekLast()] <= arr[i]) {
                indexList.pollLast();
            }
            indexList.addLast(i);
            // 当前最大值下标被窗口滑过时，弹出
            if (indexList.peekFirst() == i - w) {
                indexList.pollFirst();
            }
            // 注意是peekFirst
            if (i >= w - 1) {
                res[index++] = arr[indexList.peekFirst()];
            }
        }
        return res;
    }

    /**
     * 单调栈——无重复值
     * 获得最近小于的，栈底到栈顶是从小到大
     */
    public static int[][] getNearLessNoRepeat(int[] arr) {
        // 栈中放的是数组的下标
        Stack<Integer> stack = new Stack<>();
        // [][0]表示左边最近小于的下标，[][1]表示右边最近最小的下标
        int[][] res = new int[arr.length][2];
        for (int i = 0; i < arr.length; i++) {
            while (!stack.isEmpty() && arr[stack.peek()] > arr[i]) {
                int popIndex = stack.pop();
                // 当前下标的左边最近小于的是当前下标弹出后的栈顶
                int leftIndex = stack.isEmpty() ? -1 : stack.peek();
                res[popIndex][0] = leftIndex;
                res[popIndex][1] = i;
            }
            stack.push(i);
        }
        while (!stack.isEmpty()) {
            int popIndex = stack.pop();
            int leftIndex = stack.isEmpty() ? -1 : stack.peek();
            res[popIndex][0] = leftIndex;
            res[popIndex][1] = -1;
        }
        return res;
    }

    /**
     * 单调栈——有重复值，使用队列存储重复值下标
     * 获得最近小于的，栈底到栈顶是从小到大
     * @param arr
     * @return
     */
    public static int[][] getNearLess(int[] arr) {
        int[][] res = new int[arr.length][2];
        Stack<ArrayList<Integer>> stack = new Stack<>();
        for (int i = 0; i < arr.length; i++) {
            while (!stack.isEmpty() && arr[stack.peek().get(0)] > arr[i]) {
                ArrayList<Integer> popIndexs = stack.pop();
                // 最近的是队列中最后一个
                int leftIndex = stack.isEmpty() ? -1 : stack.peek().get(stack.peek().size() - 1);
                for (Integer index : popIndexs) {
                    res[index][0] = leftIndex;
                    res[index][1] = i;
                }
            }
            if (!stack.isEmpty() && arr[stack.peek().get(0)] == arr[i]) {
                stack.peek().add(i);
            } else {
                ArrayList<Integer> list = new ArrayList<>();
                list.add(i);
                stack.push(list);
            }
        }
        while (!stack.isEmpty()) {
            ArrayList<Integer> popIndexs = stack.pop();
            // 最近的是队列中最后一个
            int leftIndex = stack.isEmpty() ? -1 : stack.peek().get(stack.peek().size() - 1);
            for (Integer index : popIndexs) {
                res[index][0] = leftIndex;
                res[index][1] = -1;
            }
        }
        return res;
    }

    /**
     * 数组中累积和与最小值的乘积，假设叫做指标A
     * 给定一个数组，请返回子数组中，指标A最大的值。
     * 以i为最小值的子数组的范围就是左边小于i的下标的右一个，右边小于i的下标的左一个。
     * @param arr
     * @return
     */
    public static int maxA(int[] arr) {
        int[][] nearLess = getNearLess(arr);
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < arr.length; i++) {
            int sum = 0;
            int leftIndex = nearLess[i][0];
            int rightIndex = nearLess[i][1];
            if (rightIndex == -1) {
                rightIndex = arr.length;
            }
            while (leftIndex < rightIndex - 1) {
                sum = sum + arr[++leftIndex];
            }
            max = Math.max(max, sum * arr[i]);
        }
        return max;
    }


    public static void main(String[] args) {
        // 滑动窗口
        int[] arr = { 4, 3, 5, 4, 3, 3, 6, 7 };
        int w = 3;
        System.out.println(Arrays.toString(slidingWindow(arr, w)));

        // 单调栈——无重复值
        int[] arr1 = { 4, 3, 5, 6, 7 };
        int[][] nearLessNoRepeat = getNearLessNoRepeat(arr1);
        System.out.println(Arrays.deepToString(nearLessNoRepeat));

        // 单调栈——有重复值
        int[] arr2 = { 4, 3, 5, 4, 3, 3, 6, 7 };
        int[][] nearLess = getNearLess(arr2);
        System.out.println(Arrays.deepToString(nearLess));

        // 单调栈的应用
        int[] arr3 = { 4, 3, 5, 4, 3, 3, 6, 7 };
        int i = maxA(arr3);
        System.out.println(i);
    }
}

标签：peek,arr,leftIndex,int,res,算法,滑动,stack,单调
来源： https://www.cnblogs.com/shizhe99/p/16072010.html