This commit is contained in:
cezarbbb
2022-07-11 17:42:47 +08:00
11 changed files with 343 additions and 70 deletions

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@ -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:

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@ -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

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@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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<Integer, Integer> 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<Integer, Integer> 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;
}
}

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@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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;
};
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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;
}

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@ -6,6 +6,8 @@
# 143.重排链表
[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
![](https://code-thinking-1253855093.file.myqcloud.com/pics/20210726160122.png)
## 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

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@ -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;
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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;
}
```
-----------------------

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@ -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;
};
```