Merge branch 'main' of github.com:doocs/leetcode into main

Author: yanglbme

lcof/面试题17. 打印从1到最大的n位数/README.md

@@ -25,17 +25,17 @@
 ```python
 class Solution:
     def printNumbers(self, n: int) -> List[int]:
-        return [i for i in range(1, pow(10, n))]
+        return [i for i in range(1, 10 ** n)]
 ```
 
 ### **Java**
 
 ```java
 class Solution {
     public int[] printNumbers(int n) {
-        int nums = (int) Math.pow(10, n);
-        int[] res = new int[nums - 1];
-        for (int i = 0; i < nums - 1; ++i) {
+        n = (int) Math.pow(10, n) - 1;
+        int[] res = new int[n];
+        for (int i = 0; i < n; ++i) {
             res[i] = i + 1;
         }
         return res;
@@ -52,7 +52,7 @@ class Solution {
  */
 var printNumbers = function (n) {
   let res = [];
-  for (let i = 1; i < 10 ** n; i++) {
+  for (let i = 1; i < 10 ** n; ++i) {
     res.push(i);
   }
   return res;
@@ -64,12 +64,12 @@ var printNumbers = function (n) {
 ```go
 func printNumbers(n int) []int {
     d := 10
-    for i :=1; i < n ; i++ {
+    for i := 1; i < n; i++ {
         d *= 10
     }
-    res := make([]int,d-1)
-    for i:=1;i < d; i ++ {
-        res[i-1] = i
+    res := make([]int, d - 1)
+    for i := 1; i < d; i++ {
+        res[i - 1] = i
     }
     return res
 }

lcof/面试题17. 打印从1到最大的n位数/Solution.go

@@ -1,11 +1,11 @@
 func printNumbers(n int) []int {
     d := 10
-    for i :=1; i < n ; i++ {
+    for i := 1; i < n; i++ {
         d *= 10
     }
-    res := make([]int,d-1)
-    for i:=1;i < d; i ++ {
-        res[i-1] = i
+    res := make([]int, d - 1)
+    for i := 1; i < d; i++ {
+        res[i - 1] = i
     }
     return res
 }

lcof/面试题17. 打印从1到最大的n位数/Solution.java

@@ -1,8 +1,8 @@
 class Solution {
     public int[] printNumbers(int n) {
-        int nums = (int) Math.pow(10, n);
-        int[] res = new int[nums - 1];
-        for (int i = 0; i < nums - 1; ++i) {
+        n = (int) Math.pow(10, n) - 1;
+        int[] res = new int[n];
+        for (int i = 0; i < n; ++i) {
             res[i] = i + 1;
         }
         return res;

lcof/面试题17. 打印从1到最大的n位数/Solution.js

@@ -4,7 +4,7 @@
  */
 var printNumbers = function (n) {
   let res = [];
-  for (let i = 1; i < 10 ** n; i++) {
+  for (let i = 1; i < 10 ** n; ++i) {
     res.push(i);
   }
   return res;

lcof/面试题17. 打印从1到最大的n位数/Solution.py

@@ -1,3 +1,3 @@
 class Solution:
     def printNumbers(self, n: int) -> List[int]:
-        return [i for i in range(1, pow(10, n))]
+        return [i for i in range(1, 10 ** n)]

lcof/面试题28. 对称的二叉树/README.md

@@ -58,16 +58,16 @@
 
 class Solution:
     def isSymmetric(self, root: TreeNode) -> bool:
+        def is_symmetric(left, right):
+            if left is None and right is None:
+                return True
+            if left is None or right is None or left.val != right.val:
+                return False
+            return is_symmetric(left.left, right.right) and is_symmetric(left.right, right.left)
         if root is None:
             return True
-        return self.symmetric(root.left, root.right)
+        return is_symmetric(root.left, root.right)
 
-    def symmetric(self, node1, node2) -> bool:
-        if node1 is None and node2 is None:
-            return True
-        if node1 is None or node2 is None or node1.val != node2.val:
-            return False
-        return self.symmetric(node1.left, node2.right) and self.symmetric(node1.right, node2.left)
 ```
 
 ### **Java**
@@ -84,20 +84,14 @@ class Solution:
  */
 class Solution {
     public boolean isSymmetric(TreeNode root) {
-        if (root == null) {
-            return true;
-        }
+        if (root == null) return true;
         return isSymmetric(root.left, root.right);
     }
 
-    private boolean isSymmetric(TreeNode node1, TreeNode node2) {
-        if (node1 == null && node2 == null) {
-            return true;
-        }
-        if (node1 == null || node2 == null || node1.val != node2.val) {
-            return false;
-        }
-        return isSymmetric(node1.left, node2.right) && isSymmetric(node1.right, node2.left);
+    private boolean isSymmetric(TreeNode left, TreeNode right) {
+        if (left == null && right == null) return true;
+        if (left == null || right == null || left.val != right.val) return false;
+        return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
     }
 }
 ```
@@ -117,10 +111,10 @@ class Solution {
  * @return {boolean}
  */
 var isSymmetric = function (root) {
-  function dfs(a, b) {
-    if (!a && !b) return true;
-    if (!a || !b) return false;
-    return a.val === b.val && dfs(a.left, b.right) && dfs(a.right, b.left);
+  function dfs(left, right) {
+    if (!left && !right) return true;
+    if (!left || !right || left.val != right.val) return false;
+    return dfs(left.left, right.right) && dfs(left.right, right.left);
   }
   if (!root) return true;
   return dfs(root.left, root.right);
@@ -130,24 +124,29 @@ var isSymmetric = function (root) {
 ### **Go**
 
 ```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
 func isSymmetric(root *TreeNode) bool {
     if root == nil {
         return true
     }
     return isSymme(root.Left, root.Right)
 }
 
-func isSymme(a *TreeNode, b *TreeNode) bool {
-    if a == nil && b == nil {
+func isSymme(left *TreeNode, right *TreeNode) bool {
+    if left == nil && right == nil {
         return true
     }
-    if a == nil || b ==nil {
-        return false
-    }
-    if a.Val != b.Val {
+    if left == nil || right == nil || left.Val != right.Val {
         return false
     }
-    return isSymme(a.Left,b.Right) && isSymme(a.Right, b.Left)
+    return isSymme(left.Left, right.Right) && isSymme(left.Right, right.Left)
 }
 ```
 

lcof/面试题28. 对称的二叉树/Solution.go

@@ -1,19 +1,24 @@
-func isSymmetric(root *TreeNode) bool {
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+ func isSymmetric(root *TreeNode) bool {
     if root == nil {
         return true
     }
     return isSymme(root.Left, root.Right)
 }
 
-func isSymme(a *TreeNode, b *TreeNode) bool {
-    if a == nil && b == nil {
+func isSymme(left *TreeNode, right *TreeNode) bool {
+    if left == nil && right == nil {
         return true
     }
-    if a == nil || b ==nil {
+    if left == nil || right == nil || left.Val != right.Val {
         return false
     }
-    if a.Val != b.Val {
-        return false
-    }
-    return isSymme(a.Left,b.Right) && isSymme(a.Right, b.Left)
+    return isSymme(left.Left, right.Right) && isSymme(left.Right, right.Left)
 }

lcof/面试题28. 对称的二叉树/Solution.java

@@ -9,19 +9,13 @@
  */
 class Solution {
     public boolean isSymmetric(TreeNode root) {
-        if (root == null) {
-            return true;
-        }
+        if (root == null) return true;
         return isSymmetric(root.left, root.right);
     }
 
-    private boolean isSymmetric(TreeNode node1, TreeNode node2) {
-        if (node1 == null && node2 == null) {
-            return true;
-        }
-        if (node1 == null || node2 == null || node1.val != node2.val) {
-            return false;
-        }
-        return isSymmetric(node1.left, node2.right) && isSymmetric(node1.right, node2.left);
+    private boolean isSymmetric(TreeNode left, TreeNode right) {
+        if (left == null && right == null) return true;
+        if (left == null || right == null || left.val != right.val) return false;
+        return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
     }
 }

lcof/面试题28. 对称的二叉树/Solution.js

@@ -10,10 +10,10 @@
  * @return {boolean}
  */
 var isSymmetric = function (root) {
-  function dfs(a, b) {
-    if (!a && !b) return true;
-    if (!a || !b) return false;
-    return a.val === b.val && dfs(a.left, b.right) && dfs(a.right, b.left);
+  function dfs(left, right) {
+    if (!left && !right) return true;
+    if (!left || !right || left.val != right.val) return false;
+    return dfs(left.left, right.right) && dfs(left.right, right.left);
   }
   if (!root) return true;
   return dfs(root.left, root.right);

lcof/面试题28. 对称的二叉树/Solution.py

@@ -7,13 +7,12 @@
 
 class Solution:
     def isSymmetric(self, root: TreeNode) -> bool:
+        def is_symmetric(left, right):
+            if left is None and right is None:
+                return True
+            if left is None or right is None or left.val != right.val:
+                return False
+            return is_symmetric(left.left, right.right) and is_symmetric(left.right, right.left)
         if root is None:
             return True
-        return self.symmetric(root.left, root.right)
-    
-    def symmetric(self, node1, node2) -> bool:
-        if node1 is None and node2 is None:
-            return True
-        if node1 is None or node2 is None or node1.val != node2.val:
-            return False
-        return self.symmetric(node1.left, node2.right) and self.symmetric(node1.right, node2.left)
+        return is_symmetric(root.left, root.right)