Java与算法（3）

Java与算法（3）

题目：

定义二叉树节点如下： public class Node { public int value; public Node left; public Node right;

public Node(int data) {        this.value = data;    }}

一个数组的MaxTree定义如下

1 数组没有重复元素 2 MaxTree是一棵二叉树，数组的每一个值对应一个二叉树节点

3包括MaxTree树在内且其中的每一棵子树上，值 最大的节点都是树的头

给定一个没有重复元素的数组arr，写出生成这个数组的MaxTree函数，要求如果数组长度为N，则时间复杂度为O（N），额外空间复杂度为O（N）

public class MaxTree {/* * 节点类 */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<Node> stack = new Stack<Node>();HashMap<Node, Node> lBigMap = new HashMap<Node, Node>();//存放节点对应的左边第一个比它大的节点HashMap<Node, Node> rBigMap = new HashMap<Node, Node>();//存放节点对应的右边第一个比它大的节点for (int i = 0; i != nArr.length; i++) {Node curNode = nArr[i];while ((!stack.isEmpty()) && stack.peek().value < curNode.value) {popStackSetMap(stack, lBigMap);}stack.push(curNode);}while (!stack.isEmpty()) {popStackSetMap(stack, lBigMap);}for (int i = nArr.length - 1; i != -1; i--) {Node curNode = nArr[i];while ((!stack.isEmpty()) && stack.peek().value < curNode.value) {popStackSetMap(stack, rBigMap);}stack.push(curNode);}while (!stack.isEmpty()) {popStackSetMap(stack, rBigMap);}Node head = null;for (int i = 0; i != nArr.length; i++) {Node curNode = nArr[i];Node left = lBigMap.get(curNode);Node right = rBigMap.get(curNode);if (left == null && right == null) {head = curNode;} else if (left == null) {if (right.left == null) {right.left = curNode;} else {right.right = curNode;}} else if (right == null) {if (left.left == null) {left.left = curNode;} else {left.right = curNode;}} else {Node parent = left.value < right.value ? left : right;if (parent.left == null) {parent.left = curNode;} else {parent.right = curNode;}}}return head;}public static void popStackSetMap(Stack<Node> stack, HashMap<Node, Node> map) {Node popNode = stack.pop();if (stack.isEmpty()) {map.put(popNode, null);} else {map.put(popNode, stack.peek());}}/* * 先序遍历 */public static void printPreOrder(Node head) {if (head == null) {return;}System.out.print(head.value + " ");printPreOrder(head.left);printPreOrder(head.right);}/* * 中序遍历 */public static void printInOrder(Node head) {if (head == null) {return;}printPreOrder(head.left);System.out.print(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);}}

题目：

给定一个整型矩阵map，其中只有0和1两种，求其中全是1的所有矩形区域中，最大的矩形区域为1的数量

例如：

1,1,1,0

其中，最大的矩形区域有3个1，所以返回3

两种解法：

public class MaximalRectangle {public static int 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 < map.length; i++) {for (int j = 0; j < map[0].length; j++) {height[j] = map[i][j] == 0 ? 0 : height[j] + 1;}maxArea = Math.max(maxRecFromBottom(height), maxArea);}return maxArea;}public static int maxRecFromBottom(int[] height) {if (height == null || height.length == 0) {return 0;}int maxArea = 0;Stack<Integer> stack = new Stack<Integer>();for (int i = 0; i < height.length; i++) {while (!stack.isEmpty() && height[i] <= height[stack.peek()]) {int j = stack.pop();int k = stack.isEmpty() ? -1 : stack.peek();int curArea = (i - k - 1) * height[j];maxArea = Math.max(maxArea, curArea);}stack.push(i);}while (!stack.isEmpty()) {int j = stack.pop();int k = stack.isEmpty() ? -1 : stack.peek();int curArea = (height.length - k - 1) * height[j];maxArea = Math.max(maxArea, curArea);}return maxArea;}public static void main(String[] args) {int[][] map = { { 1, 0, 1, 1 }, { 1, 1, 1, 1 }, { 1, 1, 1, 0 }, };System.out.println(maxRecSize(map));}}

public class Exam {static int width,height;static int max=0;public static void dg(int sz[][],int x,int y,int w,int h){int num_1=0;for(int i=0;i<w;i++)for(int j=0;j<h;j++){if(sz[x+i][y+j]==1){num_1++;}}if((w*h)==num_1){if(num_1>max){max=num_1;}if(w+1+x<=width){dg(sz,x,y,w+1,h);}if(h+1+y<=height){dg(sz,x,y,w,h+1);}}}public static void main(String args[]){width=5;height=5;int shuzu[][]={{1,1,1,1,1},  {1,0,1,0,1},  {1,1,1,1,1},  {1,1,1,0,1},  {1,1,1,1,0}};for(int i=0;i<width;i++)for(int j=0;j<height;j++){dg(shuzu,i,j,1,1);}System.out.print(max);}}