diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md
index e636bfff..7ef86562 100644
--- a/problems/0024.两两交换链表中的节点.md
+++ b/problems/0024.两两交换链表中的节点.md
@@ -254,20 +254,19 @@ TypeScript:
```typescript
function swapPairs(head: ListNode | null): ListNode | null {
- const dummyHead: ListNode = new ListNode(0, head);
- let cur: ListNode = dummyHead;
- while(cur.next !== null && cur.next.next !== null) {
- const tem: ListNode = cur.next;
- const tem1: ListNode = cur.next.next.next;
-
- cur.next = cur.next.next; // step 1
- cur.next.next = tem; // step 2
- cur.next.next.next = tem1; // step 3
-
- cur = cur.next.next;
- }
- return dummyHead.next;
-}
+ const dummyNode: ListNode = new ListNode(0, head);
+ let curNode: ListNode | null = dummyNode;
+ while (curNode && curNode.next && curNode.next.next) {
+ let firstNode: ListNode = curNode.next,
+ secNode: ListNode = curNode.next.next,
+ thirdNode: ListNode | null = curNode.next.next.next;
+ curNode.next = secNode;
+ secNode.next = firstNode;
+ firstNode.next = thirdNode;
+ curNode = firstNode;
+ }
+ return dummyNode.next;
+};
```
Kotlin:
diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md
index ed7c49b4..77def4f8 100644
--- a/problems/0035.搜索插入位置.md
+++ b/problems/0035.搜索插入位置.md
@@ -283,6 +283,28 @@ var searchInsert = function (nums, target) {
};
```
+### TypeScript
+
+```typescript
+// 第一种二分法
+function searchInsert(nums: number[], target: number): number {
+ const length: number = nums.length;
+ let left: number = 0,
+ right: number = length - 1;
+ while (left <= right) {
+ const mid: number = Math.floor((left + right) / 2);
+ if (nums[mid] < target) {
+ left = mid + 1;
+ } else if (nums[mid] === target) {
+ return mid;
+ } else {
+ right = mid - 1;
+ }
+ }
+ return right + 1;
+};
+```
+
### Swift
```swift
diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md
index 41940487..11ca2d03 100644
--- a/problems/0093.复原IP地址.md
+++ b/problems/0093.复原IP地址.md
@@ -659,6 +659,48 @@ func restoreIpAddresses(_ s: String) -> [String] {
}
```
+## Scala
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def restoreIpAddresses(s: String): List[String] = {
+ var result = mutable.ListBuffer[String]()
+ if (s.size < 4 || s.length > 12) return result.toList
+ var path = mutable.ListBuffer[String]()
+
+ // 判断IP中的一个字段是否为正确的
+ def isIP(sub: String): Boolean = {
+ if (sub.size > 1 && sub(0) == '0') return false
+ if (sub.toInt > 255) return false
+ true
+ }
+
+ def backtracking(startIndex: Int): Unit = {
+ if (startIndex >= s.size) {
+ if (path.size == 4) {
+ result.append(path.mkString(".")) // mkString方法可以把集合里的数据以指定字符串拼接
+ return
+ }
+ return
+ }
+ // subString
+ for (i <- startIndex until startIndex + 3 if i < s.size) {
+ var subString = s.substring(startIndex, i + 1)
+ if (isIP(subString)) { // 如果合法则进行下一轮
+ path.append(subString)
+ backtracking(i + 1)
+ path = path.take(path.size - 1)
+ }
+ }
+ }
+
+ backtracking(0)
+ result.toList
+ }
+}
+```
+
-----------------------
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 6d87c756..91e0c8d8 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
@@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3], 后序遍历是[3 2 1]。
```java
class Solution {
+ Map map; // 方便根据数值查找位置
public TreeNode buildTree(int[] inorder, int[] postorder) {
- return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
+ map = new HashMap<>();
+ for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+ map.put(inorder[i], i);
+ }
+
+ return findNode(inorder, 0, inorder.length, postorder,0, postorder.length); // 前闭后开
}
- public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
- int[] postorder, int postLeft, int postRight) {
- // 没有元素了
- if (inRight - inLeft < 1) {
+
+ public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
+ // 参数里的范围都是前闭后开
+ if (inBegin >= inEnd || postBegin >= postEnd) { // 不满足左闭右开,说明没有元素,返回空树
return null;
}
- // 只有一个元素了
- if (inRight - inLeft == 1) {
- return new TreeNode(inorder[inLeft]);
- }
- // 后序数组postorder里最后一个即为根结点
- int rootVal = postorder[postRight - 1];
- TreeNode root = new TreeNode(rootVal);
- int rootIndex = 0;
- // 根据根结点的值找到该值在中序数组inorder里的位置
- for (int i = inLeft; i < inRight; i++) {
- if (inorder[i] == rootVal) {
- rootIndex = i;
- break;
- }
- }
- // 根据rootIndex划分左右子树
- root.left = buildTree1(inorder, inLeft, rootIndex,
- postorder, postLeft, postLeft + (rootIndex - inLeft));
- root.right = buildTree1(inorder, rootIndex + 1, inRight,
- postorder, postLeft + (rootIndex - inLeft), postRight - 1);
+ int rootIndex = map.get(postorder[postEnd - 1]); // 找到后序遍历的最后一个元素在中序遍历中的位置
+ TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
+ int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数,用来确定后序数列的个数
+ root.left = findNode(inorder, inBegin, rootIndex,
+ postorder, postBegin, postBegin + lenOfLeft);
+ root.right = findNode(inorder, rootIndex + 1, inEnd,
+ postorder, postBegin + lenOfLeft, postEnd - 1);
+
return root;
}
}
@@ -622,31 +616,29 @@ class Solution {
```java
class Solution {
+ Map map;
public TreeNode buildTree(int[] preorder, int[] inorder) {
- return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
- }
-
- public TreeNode helper(int[] preorder, int preLeft, int preRight,
- int[] inorder, int inLeft, int inRight) {
- // 递归终止条件
- if (inLeft > inRight || preLeft > preRight) return null;
-
- // val 为前序遍历第一个的值,也即是根节点的值
- // idx 为根据根节点的值来找中序遍历的下标
- int idx = inLeft, val = preorder[preLeft];
- TreeNode root = new TreeNode(val);
- for (int i = inLeft; i <= inRight; i++) {
- if (inorder[i] == val) {
- idx = i;
- break;
- }
+ map = new HashMap<>();
+ for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+ map.put(inorder[i], i);
}
- // 根据 idx 来递归找左右子树
- root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
- inorder, inLeft, idx - 1);
- root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
- inorder, idx + 1, inRight);
+ return findNode(preorder, 0, preorder.length, inorder, 0, inorder.length); // 前闭后开
+ }
+
+ public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) {
+ // 参数里的范围都是前闭后开
+ if (preBegin >= preEnd || inBegin >= inEnd) { // 不满足左闭右开,说明没有元素,返回空树
+ return null;
+ }
+ int rootIndex = map.get(preorder[preBegin]); // 找到前序遍历的第一个元素在中序遍历中的位置
+ TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
+ int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数,用来确定前序数列的个数
+ root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1,
+ inorder, inBegin, rootIndex);
+ root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd,
+ inorder, rootIndex + 1, inEnd);
+
return root;
}
}
diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md
index a371864d..64d45853 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -676,5 +676,50 @@ impl Solution {
}
}
```
+
+
+## Scala
+
+```scala
+object Solution {
+
+ import scala.collection.mutable
+
+ def partition(s: String): List[List[String]] = {
+ var result = mutable.ListBuffer[List[String]]()
+ var path = mutable.ListBuffer[String]()
+
+ // 判断字符串是否回文
+ def isPalindrome(start: Int, end: Int): Boolean = {
+ var (left, right) = (start, end)
+ while (left < right) {
+ if (s(left) != s(right)) return false
+ left += 1
+ right -= 1
+ }
+ true
+ }
+
+ // 回溯算法
+ def backtracking(startIndex: Int): Unit = {
+ if (startIndex >= s.size) {
+ result.append(path.toList)
+ return
+ }
+ // 添加循环守卫,如果当前分割是回文子串则进入回溯
+ for (i <- startIndex until s.size if isPalindrome(startIndex, i)) {
+ path.append(s.substring(startIndex, i + 1))
+ backtracking(i + 1)
+ path = path.take(path.size - 1)
+ }
+ }
+
+ backtracking(0)
+ result.toList
+ }
+}
+```
+
+
-----------------------
diff --git a/problems/0141.环形链表.md b/problems/0141.环形链表.md
index ddd83c94..ce90b6c4 100644
--- a/problems/0141.环形链表.md
+++ b/problems/0141.环形链表.md
@@ -7,6 +7,8 @@
# 141. 环形链表
+[力扣题目链接](https://leetcode.cn/problems/linked-list-cycle/submissions/)
+
给定一个链表,判断链表中是否有环。
如果链表中有某个节点,可以通过连续跟踪 next 指针再次到达,则链表中存在环。 为了表示给定链表中的环,我们使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始)。 如果 pos 是 -1,则在该链表中没有环。注意:pos 不作为参数进行传递,仅仅是为了标识链表的实际情况。
@@ -103,7 +105,7 @@ class Solution:
return False
```
-## Go
+### Go
```go
func hasCycle(head *ListNode) bool {
@@ -139,6 +141,23 @@ var hasCycle = function(head) {
};
```
+### TypeScript
+
+```typescript
+function hasCycle(head: ListNode | null): boolean {
+ let slowNode: ListNode | null = head,
+ fastNode: ListNode | null = head;
+ while (fastNode !== null && fastNode.next !== null) {
+ slowNode = slowNode!.next;
+ fastNode = fastNode.next.next;
+ if (slowNode === fastNode) return true;
+ }
+ return false;
+};
+```
+
+
+
-----------------------
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md
index 6b7c7e66..658bd868 100644
--- a/problems/0142.环形链表II.md
+++ b/problems/0142.环形链表II.md
@@ -301,13 +301,13 @@ function detectCycle(head: ListNode | null): ListNode | null {
let slowNode: ListNode | null = head,
fastNode: ListNode | null = head;
while (fastNode !== null && fastNode.next !== null) {
- slowNode = (slowNode as ListNode).next;
+ slowNode = slowNode!.next;
fastNode = fastNode.next.next;
if (slowNode === fastNode) {
slowNode = head;
while (slowNode !== fastNode) {
- slowNode = (slowNode as ListNode).next;
- fastNode = (fastNode as ListNode).next;
+ slowNode = slowNode!.next;
+ fastNode = fastNode!.next;
}
return slowNode;
}
diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md
index 790bcb48..c60fc0f9 100644
--- a/problems/0143.重排链表.md
+++ b/problems/0143.重排链表.md
@@ -6,6 +6,8 @@
# 143.重排链表
+[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
+

## 思路
@@ -465,7 +467,81 @@ var reorderList = function(head, s = [], tmp) {
}
```
+### TypeScript
+
+> 辅助数组法:
+
+```typescript
+function reorderList(head: ListNode | null): void {
+ if (head === null) return;
+ const helperArr: ListNode[] = [];
+ let curNode: ListNode | null = head;
+ while (curNode !== null) {
+ helperArr.push(curNode);
+ curNode = curNode.next;
+ }
+ let node: ListNode = head;
+ let left: number = 1,
+ right: number = helperArr.length - 1;
+ let count: number = 0;
+ while (left <= right) {
+ if (count % 2 === 0) {
+ node.next = helperArr[right--];
+ } else {
+ node.next = helperArr[left++];
+ }
+ count++;
+ node = node.next;
+ }
+ node.next = null;
+};
+```
+
+> 分割链表法:
+
+```typescript
+function reorderList(head: ListNode | null): void {
+ if (head === null || head.next === null) return;
+ let fastNode: ListNode = head,
+ slowNode: ListNode = head;
+ while (fastNode.next !== null && fastNode.next.next !== null) {
+ slowNode = slowNode.next!;
+ fastNode = fastNode.next.next;
+ }
+ let head1: ListNode | null = head;
+ // 反转后半部分链表
+ let head2: ListNode | null = reverseList(slowNode.next);
+ // 分割链表
+ slowNode.next = null;
+ /**
+ 直接在head1链表上进行插入
+ head1 链表长度一定大于或等于head2,
+ 因此在下面的循环中,只要head2不为null, head1 一定不为null
+ */
+ while (head2 !== null) {
+ const tempNode1: ListNode | null = head1!.next,
+ tempNode2: ListNode | null = head2.next;
+ head1!.next = head2;
+ head2.next = tempNode1;
+ head1 = tempNode1;
+ head2 = tempNode2;
+ }
+};
+function reverseList(head: ListNode | null): ListNode | null {
+ let curNode: ListNode | null = head,
+ preNode: ListNode | null = null;
+ while (curNode !== null) {
+ const tempNode: ListNode | null = curNode.next;
+ curNode.next = preNode;
+ preNode = curNode;
+ curNode = tempNode;
+ }
+ return preNode;
+}
+```
+
### C
+
方法三:反转链表
```c
//翻转链表
diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md
index e2eb378e..090e73f0 100644
--- a/problems/0209.长度最小的子数组.md
+++ b/problems/0209.长度最小的子数组.md
@@ -465,6 +465,27 @@ object Solution {
}
}
```
-
+C#:
+```csharp
+public class Solution {
+ public int MinSubArrayLen(int s, int[] nums) {
+ int n = nums.Length;
+ int ans = int.MaxValue;
+ int start = 0, end = 0;
+ int sum = 0;
+ while (end < n) {
+ sum += nums[end];
+ while (sum >= s)
+ {
+ ans = Math.Min(ans, end - start + 1);
+ sum -= nums[start];
+ start++;
+ }
+ end++;
+ }
+ return ans == int.MaxValue ? 0 : ans;
+ }
+}
+```
-----------------------
diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md
index f591fcef..eeee6fa5 100644
--- a/problems/0234.回文链表.md
+++ b/problems/0234.回文链表.md
@@ -273,7 +273,7 @@ class Solution:
return pre
```
-## Go
+### Go
```go
@@ -319,6 +319,63 @@ var isPalindrome = function(head) {
};
```
+### TypeScript
+
+> 数组模拟
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+ const helperArr: number[] = [];
+ let curNode: ListNode | null = head;
+ while (curNode !== null) {
+ helperArr.push(curNode.val);
+ curNode = curNode.next;
+ }
+ let left: number = 0,
+ right: number = helperArr.length - 1;
+ while (left < right) {
+ if (helperArr[left++] !== helperArr[right--]) return false;
+ }
+ return true;
+};
+```
+
+> 反转后半部分链表
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+ if (head === null || head.next === null) return true;
+ let fastNode: ListNode | null = head,
+ slowNode: ListNode = head,
+ preNode: ListNode = head;
+ while (fastNode !== null && fastNode.next !== null) {
+ preNode = slowNode;
+ slowNode = slowNode.next!;
+ fastNode = fastNode.next.next;
+ }
+ preNode.next = null;
+ let cur1: ListNode | null = head;
+ let cur2: ListNode | null = reverseList(slowNode);
+ while (cur1 !== null) {
+ if (cur1.val !== cur2!.val) return false;
+ cur1 = cur1.next;
+ cur2 = cur2!.next;
+ }
+ return true;
+};
+function reverseList(head: ListNode | null): ListNode | null {
+ let curNode: ListNode | null = head,
+ preNode: ListNode | null = null;
+ while (curNode !== null) {
+ let tempNode: ListNode | null = curNode.next;
+ curNode.next = preNode;
+ preNode = curNode;
+ curNode = tempNode;
+ }
+ return preNode;
+}
+```
+
-----------------------
diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md
index 3138f60f..031bc0ad 100644
--- a/problems/0344.反转字符串.md
+++ b/problems/0344.反转字符串.md
@@ -190,13 +190,13 @@ javaScript:
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function(s) {
- return s.reverse();
+ //Do not return anything, modify s in-place instead.
+ reverse(s)
};
-var reverseString = function(s) {
+var reverse = function(s) {
let l = -1, r = s.length;
while(++l < --r) [s[l], s[r]] = [s[r], s[l]];
- return s;
};
```