单调栈的应用
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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。
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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
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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
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Image 4.png
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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)
-
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);
}
}
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