算法进阶三

单调栈的应用

Image 14.png
Image 15.png

单调栈的做法:找到每个数左边第一个比它大的数,右边第一个比它大的数串到它下面。

  • 证明 :形成的不是森林,而是一个颗数目。
    首先,数组中没有重复值。最大值一定会作为整棵树的头结点。任何一个节点都会找一个比他大的窜到他底下。所以,每一个节点都有归属,最终以最大值作为头部。因此,是一个树,不是多颗树,形不成森林。

  • 证明:这个流程的正确性,不会形成多叉树,最多形成二叉树。
    因为我们的逻辑是:左边离我最近比我大的数,右边离我最近的比我大的数,挂在这两个数中较小的那个数的下面。会不会产生一个孩子有多个节点的时候。

Image 16.png

package com.znst;

import java.util.HashMap;
import java.util.LinkedList;
import java.util.Stack;

public class Demo2 {
    
    public static class Node{
        public int value;
        public Node left;
        public Node right;
        
        public Node(int data) {
            this.value = data;
        }
        
        public static Node getMaxTree(int[] arr) {
            Node[] nArr = new Node[arr.length];
            for(int i=0;i!=arr.length;i++) {
                nArr[i]=new Node(arr[i]);
            }
            Stack stack = new Stack();
            HashMap lBitmap = new HashMap();
            HashMap rBitmap = new HashMap();
            for(int i=0;i!=nArr.length;i++) {
                Node curNode = nArr[i];
                while((!stack.isEmpty())&&stack.peek().value stack,HashMap map) {
        Node popNode = stack.pop();
        if(stack.isEmpty()) {
            map.put(popNode, null);
        }else {
            map.put(popNode, stack.peek());
        }
    }
    public static void printPreOrder(Node head) {
        if(head == null) {
            return ;
        }
        System.out.print(head.value+" ");
        printPreOrder(head.left);
        printPreOrder(head.right);
    }
    public static void printInOrder(Node head) {
        if(head==null) {
            return;
        }
        printPreOrder(head.left);
        System.out.println(head.value+" ");
        printPreOrder(head.right);
    }
    
    public static void main(String[] args) {
        int[] uniqueArr = {3,4,5,1,2};
        Node head = getMaxTree(uniqueArr);
        printPreOrder(head);
        System.out.println();
        printInOrder(head);
    }
}

求最大子矩阵大小

Image 17.png

Image 19.png
package com.znst;

import java.util.Stack;

public class Demo3 {

    public static maxRecSize(int[][] map) {
        if(map==null || map.length=0||map[0].length==0) {
            return 0;
        }
        int maxArea =0;
        int[] height = new int[map[0].length];
        for(int i=0;i stack = new Stack();
        for(int i=0 ;i
案例:
Image 20.png

Image 1.png

证明:
思想:用小的去找大的,最小的找到第一大的就停,所以在最高和次高中间,从i出发,一定找到2个比他大的


Image 22.png

Image 2.png

Image 3.png

Image 4.png

Image 5.png

Image 6.png
package com.znst;

import java.util.Scanner;
import java.util.Stack;

public class Demo4 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        while(in.hasNextInt()) {
            int size = in.nextInt();
            int[] arr = new int[size];
            for(int i=0;i stack = new Stack();
        stack.push(new Pair(value));
        while(index!=maxIndex) {
            value = arr[index];
            while(!stack.isEmpty()&&stack.peek().value1) {
                    res+=times;
                }else {
                    res+=stack.peek().times>1?times:0;
                }
            }
        }
        return res;
    }
    
}

Morris遍历:利用Morris遍历实现二叉树的先序,中序,后序遍历,时间复杂度O(N),额外空间复杂度O(1)。

来到的当前节点,记为Cur(引用)
1)如果cur无左孩子,cur向右移动(cur = cur.right)

  1. 如果cur有左孩子,找到cur左子树上最右的节点,记为mostright
    a.如果mostright的right指针指向空,让其指向cur,cur向左移动(cur=cur.left)
    b.如果mostright指向cur,让其指向空,cur向右移动


    Image 7.png
Image 8.png

当cur来到4节点时,4的指针指向2,


Image 9.png

Image 10.png

Image 11.png
package com.znst;

import java.util.Scanner;
import java.util.Stack;

public class Demo4 {

    public static void main(String[] args) {
    
    }
    
    public static void process(Node head) {
        if(head == null) {
            return;
        }
        //1
        System.out.println(head.value);
        process(head.left);
        //2
        System.out.println(head.value);
        process(head.right);
        //3
        System.out.println(head.value);
    }
    
    public static class Node{
        public int value;
        Node left;
        Node right;
        public Node(int data) {
            this.value = data;
        }
    }

    public static void morrisIn(Node head) {
        if(head ==null) {
            return ;
        }
        Node cur = head;
        Node mostRight = null;
        while(cur!=null) {
            mostRight = cur.left;
            if(mostRight!=null) {//左孩子不为空
                while(mostRight.right!=null&&mostRight.right!=cur) {
                    mostRight = mostRight.right;
                }
                if(mostRight.right == null) {
                    mostRight.right = cur;
                    cur = cur.left;
                    continue;
                }else { 
                    mostRight.right = null;
                }
            }
            System.out.print(cur.value+" ");
            cur = cur.right;
        }
        System.out.println();
    }
    
    /*
     * morris改先序遍历
     */
    public static void morrisPre(Node head) {
        if(head==null) {
            return;
        }
        Node cur = head;
        Node mostRight = null;
        while(cur!=null) {
            mostRight = cur.left;
            if(mostRight!=null) {
                while(mostRight.right!=null&&mostRight.right!=cur) {
                    mostRight = mostRight.right;
                }
                if(mostRight.right ==null) {
                    mostRight.right = cur;
                    System.out.println(cur.value+" ");
                    cur = cur.left;
                    continue;
                }else {
                    mostRight.right = null;
                }
            }else {//当前节点没有左子树
                System.out.print(cur.value+" ");
            }
            cur = cur.right;
        }
        System.out.println();
    }
    
    public static void morrisPos(Node head) {
        if(head == null) {
            return ;
        }
        Node cur1 = head;
        Node cur2 = null;
        while(cur1!=null) {
            cur2 = cur1.left;
            if(cur2!=null) {
                while(cur2.right!=null&&cur2.right!=cur1) {
                    cur2 = cur2.right;
                }
                if(cur2.right ==null) {
                    cur2.right = cur1;
                    cur1 = cur1.left;
                    continue;
                }else {
                    cur2.right = null;
                    printEdge(cur1.left);
                }
            }
            cur1 = cur1.left;
        }
        printEdge(head); 
        System.out.println();
    }
    
    public static void printEdge(Node head) {
        Node tail = reverseEdge(head);
        Node cur = tail;
        while(cur != null) {
            System.out.print(cur.value+" ");
            cur = cur.right;
        }
        reverseEdge(tail);
    }
}

Image 12.png

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