Java十大经典排序算法的实现图解
343
2022-08-19
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
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List
val = _val;
children = _children;
}
};
*/
class Solution {
ArrayList
public List
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
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List
val = _val;
children = _children;
}
};
*/
class Solution {
ArrayList
public List
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
public Node() {}
public Node(int _val) {
val = _val;
}
public Node(int _val, List
val = _val;
children = _children;
}
};
*/
class Solution {
ArrayList
public List
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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