Merge branch 'youngyangyang04:master' into leetcode-modify-the-code-of-the-greedy

This commit is contained in:
Yuhao Ju
2022-12-17 01:30:40 +08:00
committed by GitHub
2 changed files with 94 additions and 23 deletions

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@ -1149,6 +1149,52 @@ object Solution {
}
```
## rust
106 从中序与后序遍历序列构造二叉树
```rust
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn build_tree(inorder: Vec<i32>, postorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
if inorder.is_empty() {
return None;
}
let mut postorder = postorder;
let root = postorder.pop().unwrap();
let index = inorder.iter().position(|&x| x == root).unwrap();
let mut root = TreeNode::new(root);
root.left = Self::build_tree(inorder[..index].to_vec(), postorder[..index].to_vec());
root.right = Self::build_tree(inorder[index + 1..].to_vec(), postorder[index..].to_vec());
Some(Rc::new(RefCell::new(root)))
}
}
```
105 从前序与中序遍历序列构造二叉树
```rust
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn build_tree(preorder: Vec<i32>, inorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
if preorder.is_empty() {
return None;
}
let root = preorder[0];
let index = inorder.iter().position(|&x| x == root).unwrap();
let mut root = TreeNode::new(root);
root.left = Self::build_tree(preorder[1..index + 1].to_vec(), inorder[0..index].to_vec());
root.right = Self::build_tree(
preorder[index + 1..].to_vec(),
inorder[index + 1..].to_vec(),
);
Some(Rc::new(RefCell::new(root)))
}
}
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

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@ -271,39 +271,64 @@ int main() {
### java
```java
public class BagProblem {
public static void main(String[] args) {
int[] weight = {1, 3, 4};
int[] value = {15, 20, 30};
int bagsize = 4;
testweightbagproblem(weight, value, bagsize);
int[] weight = {1,3,4};
int[] value = {15,20,30};
int bagSize = 4;
testWeightBagProblem(weight,value,bagSize);
}
public static void testweightbagproblem(int[] weight, int[] value, int bagsize){
int wlen = weight.length, value0 = 0;
//定义dp数组dp[i][j]表示背包容量为j时前i个物品能获得的最大价值
int[][] dp = new int[wlen + 1][bagsize + 1];
//初始化:背包容量为0时能获得的价值都为0
for (int i = 0; i <= wlen; i++){
dp[i][0] = value0;
/**
* 动态规划获得结果
* @param weight 物品的重量
* @param value 物品的价值
* @param bagSize 背包容量
*/
public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
// 创建dp数组
int goods = weight.length; // 获取物品的数量
int[][] dp = new int[goods][bagSize + 1];
// 初始化dp数组
// 创建数组后其中默认的值就是0
for (int j = weight[0]; j <= bagSize; j++) {
dp[0][j] = value[0];
}
//遍历顺序:先遍历物品,再遍历背包容量
for (int i = 1; i <= wlen; i++){
for (int j = 1; j <= bagsize; j++){
if (j < weight[i - 1]){
dp[i][j] = dp[i - 1][j];
}else{
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
// 填充dp数组
for (int i = 1; i < weight.length; i++) {
for (int j = 1; j <= bagSize; j++) {
if (j < weight[i]) {
/**
* 当前背包的容量都没有当前物品i大的时候是不放物品i的
* 那么前i-1个物品能放下的最大价值就是当前情况的最大价值
*/
dp[i][j] = dp[i-1][j];
} else {
/**
* 当前背包的容量可以放下物品i
* 那么此时分两种情况:
* 1、不放物品i
* 2、放物品i
* 比较这两种情况下,哪种背包中物品的最大价值最大
*/
dp[i][j] = Math.max(dp[i-1][j] , dp[i-1][j-weight[i]] + value[i]);
}
}
}
//打印dp数组
for (int i = 0; i <= wlen; i++){
for (int j = 0; j <= bagsize; j++){
System.out.print(dp[i][j] + " ");
// 打印dp数组
for (int i = 0; i < goods; i++) {
for (int j = 0; j <= bagSize; j++) {
System.out.print(dp[i][j] + "\t");
}
System.out.print("\n");
System.out.println("\n");
}
}
}
```
### python