网页视频下载快捷键,关键词优化seo外包,160mk2成色,天津网站备案在哪照相文章目录 一、重建二叉树二、数值的整数次方三、打印从 1 到最大的 n 位数四、二叉搜索树的后序遍历序列五、数组中的逆序对 一、重建二叉树 public class Solution {int[] preorder;HashMapInteger, Integer dic new HashMap();public TreeNode buildTree(in… 文章目录 一、重建二叉树二、数值的整数次方三、打印从 1 到最大的 n 位数四、二叉搜索树的后序遍历序列五、数组中的逆序对 一、重建二叉树 public class Solution {int[] preorder;HashMapInteger, Integer dic new HashMap();public TreeNode buildTree(int[] preorder, int[] inorder) {this.preorder preorder;for (int i 0; i inorder.length; i) {dic.put(inorder[i], i);}return recur(0, 0, inorder.length - 1);}TreeNode recur(int root, int left, int right) {if (left right) {// 递归终止return null;}// 建立根节点TreeNode node new TreeNode(preorder[root]);// 划分根节点、左子树、右子树int i dic.get(preorder[root]);// 开启左子树递归node.left recur(root 1, left, i - 1);// 开启右子树递归 i - left root 1 含义为 根节点索引 左子树长度 1node.right recur(root i - left 1, i 1, right);// 回溯返回根节点return node;}public class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val x;}}}二、数值的整数次方 public class Solution {public double myPow(double x, int n) {long b n;double res 1.0;if (b 0) {x 1 / x;b -b;}while (b 0) {if ((b 1) 1) {res * x;}x * x;b 1;}return res;}
}三、打印从 1 到最大的 n 位数 public class Solution {public int[] printNumbers(int n) {int[] res new int[(int) Math.pow(10, n) - 1];for (int i 0; i res.length; i) {res[i] i 1;}return res;}
}四、二叉搜索树的后序遍历序列 public class Solution {public boolean verifyPostorder(int[] postorder) {StackInteger stack new Stack();int root Integer.MAX_VALUE;for(int i postorder.length - 1; i 0; i--) {if(postorder[i] root) {return false;}while(!stack.isEmpty() stack.peek() postorder[i]) {root stack.pop();}stack.add(postorder[i]);}return true;}
}五、数组中的逆序对 public class Solution {int[] nums, tmp;public int reversePairs(int[] nums) {this.nums nums;tmp new int[nums.length];return mergeSort(0, nums.length - 1);}private int mergeSort(int l, int r) {// 终止条件if (l r) {return 0;}// 递归划分int m (l r) / 2;int res mergeSort(l, m) mergeSort(m 1, r);// 合并阶段int i l, j m 1;for (int k l; k r; k) {tmp[k] nums[k];}for (int k l; k r; k) {if (i m 1) {nums[k] tmp[j];} else if (j r 1 || tmp[i] tmp[j])nums[k] tmp[i];else {nums[k] tmp[j];res m - i 1; // 统计逆序对}}return res;}
}
