diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 94eb405b..f8109f85 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -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, postorder: Vec) -> Option>> { + 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, inorder: Vec) -> Option>> { + 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))) + } +} +``` +

diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index ee19e53d..8f4ca2d4 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -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 背包的容量 + */ + 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