剑指Offer之Java算法习题精讲N叉树的遍历及数组与字符串

剑指Offer之Java算法习题精讲N叉树的遍历及数组与字符串

题目一

解法

/*

// Definition for a Node.

class Node {

public int val;

public List children;

public Node() {}

public Node(int _val) {

val = _val;

}

public Node(int _val, List _children) {

val = _val;

children = _children;

}

};

*/

class Solution {

ArrayList list = new ArrayList();

public List preorder(Node root) {

if (root == null) return list;

list.add(root.val);

for(int i=0;i

preorder(root.children.get(i));

}

return list;

}

}

题目二

解法

/*

// Definition for a Node.

class Node {

public int val;

public List children;

public Node() {}

public Node(int _val) {

val = _val;

}

public Node(int _val, List _children) {

val = _val;

children = _children;

}

};

*/

class Solution {

ArrayList list = new ArrayList();

public List postorder(Node root) {

if(root==null) return list;

for(int i = 0;i

postorder(root.children.get(i));

}

list.add(root.val);

return list;

}

}

题目三

解法

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode() {}

* TreeNode(int val) { this.val = val; }

* TreeNode(int val, TreeNode lefhttp://t, TreeNode right) {

* this.val = val;

* this.left = left;

* this.right = right;

* }

* }

*/

class Solution {http://

public String tree2str(TreeNode root) {

zuWmzaUl // root为空返回""

if(root==null){

return "";

}

// root左右为空

if(root.left==null&&root.right==null){

return root.val+"";

}

// root右为空

if(root.right==null){

return root.val+"("+tree2str(root.left)+")";

}

// root左右不空或是左不为空

return root.val+"("+tree2str(root.left)+")"+"("+tree2str(root.right)+")";

}

}

题目四

zuWmzaUl

解法

class Solution {

public int maximumProduct(int[] nums) {

Arrays.sort(nums);

int n = nums.length;

// 拿到最大三个数，或者最小两个数和最大一个数

int ans = Math.max(nums[n-1]*nums[n-2]*nums[n-3],nums[n-1]*nums[0]*nums[1]);

return ans;

}

}

// 拿到最大三个数，或者最小两个数和最大一个数

class Solution {

public int maximumProduct(int[] nums) {

// 最小的和第二小的

int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;

// 最大的、第二大的和第三大的

int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;

for(int n:nums){

if(n

min2 = min1;

min1 = n;

}else if(n

min2 = n;

}

if(n>max1){

max3 = max2;

max2 = max1;

max1 = n;

}else if(n>max2){

max3 = max2;

max2 = n;

}else if(n>max3){

max3 = n;

}

}

return Math.max(min1*min2*max1,max1*max2*max3);

}

}

preorder(root.children.get(i));

}

return list;

}

}

题目二

解法

/*

// Definition for a Node.

class Node {

public int val;

public List children;

public Node() {}

public Node(int _val) {

val = _val;

}

public Node(int _val, List _children) {

val = _val;

children = _children;

}

};

*/

class Solution {

ArrayList list = new ArrayList();

public List postorder(Node root) {

if(root==null) return list;

for(int i = 0;i

postorder(root.children.get(i));

}

list.add(root.val);

return list;

}

}

题目三

解法

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode() {}

* TreeNode(int val) { this.val = val; }

* TreeNode(int val, TreeNode lefhttp://t, TreeNode right) {

* this.val = val;

* this.left = left;

* this.right = right;

* }

* }

*/

class Solution {http://

public String tree2str(TreeNode root) {

zuWmzaUl // root为空返回""

if(root==null){

return "";

}

// root左右为空

if(root.left==null&&root.right==null){

return root.val+"";

}

// root右为空

if(root.right==null){

return root.val+"("+tree2str(root.left)+")";

}

// root左右不空或是左不为空

return root.val+"("+tree2str(root.left)+")"+"("+tree2str(root.right)+")";

}

}

题目四

zuWmzaUl

解法

class Solution {

public int maximumProduct(int[] nums) {

Arrays.sort(nums);

int n = nums.length;

// 拿到最大三个数，或者最小两个数和最大一个数

int ans = Math.max(nums[n-1]*nums[n-2]*nums[n-3],nums[n-1]*nums[0]*nums[1]);

return ans;

}

}

// 拿到最大三个数，或者最小两个数和最大一个数

class Solution {

public int maximumProduct(int[] nums) {

// 最小的和第二小的

int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;

// 最大的、第二大的和第三大的

int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;

for(int n:nums){

if(n

min2 = min1;

min1 = n;

}else if(n

min2 = n;

}

if(n>max1){

max3 = max2;

max2 = max1;

max1 = n;

}else if(n>max2){

max3 = max2;

max2 = n;

}else if(n>max3){

max3 = n;

}

}

return Math.max(min1*min2*max1,max1*max2*max3);

}

}

postorder(root.children.get(i));

}

list.add(root.val);

return list;

}

}

题目三

解法

/**

* Definition for a binary tree node.

* public class TreeNode {

* int val;

* TreeNode left;

* TreeNode right;

* TreeNode() {}

* TreeNode(int val) { this.val = val; }

* TreeNode(int val, TreeNode lefhttp://t, TreeNode right) {

* this.val = val;

* this.left = left;

* this.right = right;

* }

* }

*/

class Solution {http://

public String tree2str(TreeNode root) {

zuWmzaUl // root为空返回""

if(root==null){

return "";

}

// root左右为空

if(root.left==null&&root.right==null){

return root.val+"";

}

// root右为空

if(root.right==null){

return root.val+"("+tree2str(root.left)+")";

}

// root左右不空或是左不为空

return root.val+"("+tree2str(root.left)+")"+"("+tree2str(root.right)+")";

}

}

题目四

zuWmzaUl

解法

class Solution {

public int maximumProduct(int[] nums) {

Arrays.sort(nums);

int n = nums.length;

// 拿到最大三个数，或者最小两个数和最大一个数

int ans = Math.max(nums[n-1]*nums[n-2]*nums[n-3],nums[n-1]*nums[0]*nums[1]);

return ans;

}

}

// 拿到最大三个数，或者最小两个数和最大一个数

class Solution {

public int maximumProduct(int[] nums) {

// 最小的和第二小的

int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;

// 最大的、第二大的和第三大的

int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;

for(int n:nums){

if(n

min2 = min1;

min1 = n;

}else if(n

min2 = n;

}

if(n>max1){

max3 = max2;

max2 = max1;

max1 = n;

}else if(n>max2){

max3 = max2;

max2 = n;

}else if(n>max3){

max3 = n;

}

}

return Math.max(min1*min2*max1,max1*max2*max3);

}

}

min2 = min1;

min1 = n;

}else if(n

min2 = n;

}

if(n>max1){

max3 = max2;

max2 = max1;

max1 = n;

}else if(n>max2){

max3 = max2;

max2 = n;

}else if(n>max3){

max3 = n;

}

}

return Math.max(min1*min2*max1,max1*max2*max3);

}

}

min2 = n;

}

if(n>max1){

max3 = max2;

max2 = max1;

max1 = n;

}else if(n>max2){

max3 = max2;

max2 = n;

}else if(n>max3){

max3 = n;

}

}

return Math.max(min1*min2*max1,max1*max2*max3);

}

}

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