mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-11 21:10:58 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
@ -557,6 +557,37 @@ func letterCombinations(_ digits: String) -> [String] {
|
||||
}
|
||||
```
|
||||
|
||||
## Scala:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable
|
||||
def letterCombinations(digits: String): List[String] = {
|
||||
var result = mutable.ListBuffer[String]()
|
||||
if(digits == "") return result.toList // 如果参数为空,返回空结果集的List形式
|
||||
var path = mutable.ListBuffer[Char]()
|
||||
// 数字和字符的映射关系
|
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val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz")
|
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|
||||
def backtracking(index: Int): Unit = {
|
||||
if (index == digits.size) {
|
||||
result.append(path.mkString) // mkString语法:将数组类型直接转换为字符串
|
||||
return
|
||||
}
|
||||
var digit = digits(index) - '0' // 这里使用toInt会报错!必须 -'0'
|
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for (i <- 0 until map(digit).size) {
|
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path.append(map(digit)(i))
|
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backtracking(index + 1)
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||||
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>
|
||||
|
@ -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:
|
||||
|
@ -339,7 +339,6 @@ int removeElement(int* nums, int numsSize, int val){
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
Kotlin:
|
||||
```kotlin
|
||||
fun removeElement(nums: IntArray, `val`: Int): Int {
|
||||
@ -351,7 +350,6 @@ fun removeElement(nums: IntArray, `val`: Int): Int {
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
Scala:
|
||||
```scala
|
||||
object Solution {
|
||||
@ -368,5 +366,20 @@ object Solution {
|
||||
}
|
||||
```
|
||||
|
||||
C#:
|
||||
```csharp
|
||||
public class Solution {
|
||||
public int RemoveElement(int[] nums, int val) {
|
||||
int slow = 0;
|
||||
for (int fast = 0; fast < nums.Length; fast++) {
|
||||
if (val != nums[fast]) {
|
||||
nums[slow++] = nums[fast];
|
||||
}
|
||||
}
|
||||
return slow;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -480,6 +480,62 @@ var searchRange = function(nums, target) {
|
||||
return [-1, -1];
|
||||
};
|
||||
```
|
||||
|
||||
|
||||
### TypeScript
|
||||
|
||||
```typescript
|
||||
function searchRange(nums: number[], target: number): number[] {
|
||||
const leftBoard: number = getLeftBorder(nums, target);
|
||||
const rightBoard: number = getRightBorder(nums, target);
|
||||
// target 在nums区间左侧或右侧
|
||||
if (leftBoard === (nums.length - 1) || rightBoard === 0) return [-1, -1];
|
||||
// target 不存在与nums范围内
|
||||
if (rightBoard - leftBoard <= 1) return [-1, -1];
|
||||
// target 存在于nums范围内
|
||||
return [leftBoard + 1, rightBoard - 1];
|
||||
};
|
||||
// 查找第一个大于target的元素下标
|
||||
function getRightBorder(nums: number[], target: number): number {
|
||||
let left: number = 0,
|
||||
right: number = nums.length - 1;
|
||||
// 0表示target在nums区间的左边
|
||||
let rightBoard: number = 0;
|
||||
while (left <= right) {
|
||||
let mid = Math.floor((left + right) / 2);
|
||||
if (nums[mid] <= target) {
|
||||
// 右边界一定在mid右边(不含mid)
|
||||
left = mid + 1;
|
||||
rightBoard = left;
|
||||
} else {
|
||||
// 右边界在mid左边(含mid)
|
||||
right = mid - 1;
|
||||
}
|
||||
}
|
||||
return rightBoard;
|
||||
}
|
||||
// 查找第一个小于target的元素下标
|
||||
function getLeftBorder(nums: number[], target: number): number {
|
||||
let left: number = 0,
|
||||
right: number = nums.length - 1;
|
||||
// length-1表示target在nums区间的右边
|
||||
let leftBoard: number = nums.length - 1;
|
||||
while (left <= right) {
|
||||
let mid = Math.floor((left + right) / 2);
|
||||
if (nums[mid] >= target) {
|
||||
// 左边界一定在mid左边(不含mid)
|
||||
right = mid - 1;
|
||||
leftBoard = right;
|
||||
} else {
|
||||
// 左边界在mid右边(含mid)
|
||||
left = mid + 1;
|
||||
}
|
||||
}
|
||||
return leftBoard;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
### Scala
|
||||
```scala
|
||||
object Solution {
|
||||
@ -527,5 +583,6 @@ object Solution {
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -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
|
||||
|
@ -502,5 +502,35 @@ func combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable
|
||||
def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = {
|
||||
var result = mutable.ListBuffer[List[Int]]()
|
||||
var path = mutable.ListBuffer[Int]()
|
||||
|
||||
def backtracking(sum: Int, index: Int): Unit = {
|
||||
if (sum == target) {
|
||||
result.append(path.toList) // 如果正好等于target,就添加到结果集
|
||||
return
|
||||
}
|
||||
// 应该是从当前索引开始的,而不是从0
|
||||
// 剪枝优化:添加循环守卫,当sum + c(i) <= target的时候才循环,才可以进入下一次递归
|
||||
for (i <- index until candidates.size if sum + candidates(i) <= target) {
|
||||
path.append(candidates(i))
|
||||
backtracking(sum + candidates(i), i)
|
||||
path = path.take(path.size - 1)
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(0, 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>
|
||||
|
@ -693,5 +693,37 @@ func combinationSum2(_ candidates: [Int], _ target: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable
|
||||
def combinationSum2(candidates: Array[Int], target: Int): List[List[Int]] = {
|
||||
var res = mutable.ListBuffer[List[Int]]()
|
||||
var path = mutable.ListBuffer[Int]()
|
||||
var candidate = candidates.sorted
|
||||
|
||||
def backtracking(sum: Int, startIndex: Int): Unit = {
|
||||
if (sum == target) {
|
||||
res.append(path.toList)
|
||||
return
|
||||
}
|
||||
|
||||
for (i <- startIndex until candidate.size if sum + candidate(i) <= target) {
|
||||
if (!(i > startIndex && candidate(i) == candidate(i - 1))) {
|
||||
path.append(candidate(i))
|
||||
backtracking(sum + candidate(i), i + 1)
|
||||
path = path.take(path.size - 1)
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(0, 0)
|
||||
res.toList
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -673,5 +673,63 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
### Scala
|
||||
|
||||
暴力:
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable // 导包
|
||||
def combine(n: Int, k: Int): List[List[Int]] = {
|
||||
var result = mutable.ListBuffer[List[Int]]() // 存放结果集
|
||||
var path = mutable.ListBuffer[Int]() //存放符合条件的结果
|
||||
|
||||
def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
|
||||
if (path.size == k) {
|
||||
// 如果path的size == k就达到题目要求,添加到结果集,并返回
|
||||
result.append(path.toList)
|
||||
return
|
||||
}
|
||||
for (i <- startIndex to n) { // 遍历从startIndex到n
|
||||
path.append(i) // 先把数字添加进去
|
||||
backtracking(n, k, i + 1) // 进行下一步回溯
|
||||
path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(n, k, 1) // 执行回溯
|
||||
result.toList // 最终返回result的List形式,return关键字可以省略
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
剪枝:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable // 导包
|
||||
def combine(n: Int, k: Int): List[List[Int]] = {
|
||||
var result = mutable.ListBuffer[List[Int]]() // 存放结果集
|
||||
var path = mutable.ListBuffer[Int]() //存放符合条件的结果
|
||||
|
||||
def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
|
||||
if (path.size == k) {
|
||||
// 如果path的size == k就达到题目要求,添加到结果集,并返回
|
||||
result.append(path.toList)
|
||||
return
|
||||
}
|
||||
// 剪枝优化
|
||||
for (i <- startIndex to (n - (k - path.size) + 1)) {
|
||||
path.append(i) // 先把数字添加进去
|
||||
backtracking(n, k, i + 1) // 进行下一步回溯
|
||||
path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(n, k, 1) // 执行回溯
|
||||
result.toList // 最终返回result的List形式,return关键字可以省略
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -346,5 +346,34 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
Scala:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable // 导包
|
||||
def combine(n: Int, k: Int): List[List[Int]] = {
|
||||
var result = mutable.ListBuffer[List[Int]]() // 存放结果集
|
||||
var path = mutable.ListBuffer[Int]() //存放符合条件的结果
|
||||
|
||||
def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
|
||||
if (path.size == k) {
|
||||
// 如果path的size == k就达到题目要求,添加到结果集,并返回
|
||||
result.append(path.toList)
|
||||
return
|
||||
}
|
||||
// 剪枝优化
|
||||
for (i <- startIndex to (n - (k - path.size) + 1)) {
|
||||
path.append(i) // 先把数字添加进去
|
||||
backtracking(n, k, i + 1) // 进行下一步回溯
|
||||
path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(n, k, 1) // 执行回溯
|
||||
result.toList // 最终返回result的List形式,return关键字可以省略
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -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>
|
||||
|
@ -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;
|
||||
}
|
||||
}
|
||||
|
@ -448,5 +448,27 @@ struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
|
||||
}
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
递归:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def sortedArrayToBST(nums: Array[Int]): TreeNode = {
|
||||
def buildTree(left: Int, right: Int): TreeNode = {
|
||||
if (left > right) return null // 当left大于right的时候,返回空
|
||||
// 最中间的节点是当前节点
|
||||
var mid = left + (right - left) / 2
|
||||
var curNode = new TreeNode(nums(mid))
|
||||
curNode.left = buildTree(left, mid - 1)
|
||||
curNode.right = buildTree(mid + 1, right)
|
||||
curNode
|
||||
}
|
||||
buildTree(0, nums.size - 1)
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -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>
|
||||
|
@ -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>
|
||||
|
@ -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;
|
||||
}
|
||||
|
@ -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
|
||||
|
@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int {
|
||||
return stack.last!
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
PHP:
|
||||
```php
|
||||
class Solution {
|
||||
function evalRPN($tokens) {
|
||||
$st = new SplStack();
|
||||
for($i = 0;$i<count($tokens);$i++){
|
||||
// 是数字直接入栈
|
||||
if(is_numeric($tokens[$i])){
|
||||
$st->push($tokens[$i]);
|
||||
}else{
|
||||
// 是符号进行运算
|
||||
$num1 = $st->pop();
|
||||
$num2 = $st->pop();
|
||||
if ($tokens[$i] == "+") $st->push($num2 + $num1);
|
||||
if ($tokens[$i] == "-") $st->push($num2 - $num1);
|
||||
if ($tokens[$i] == "*") $st->push($num2 * $num1);
|
||||
// 注意处理小数部分
|
||||
if ($tokens[$i] == "/") $st->push(intval($num2 / $num1));
|
||||
}
|
||||
}
|
||||
return $st->pop();
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
Scala:
|
||||
```scala
|
||||
object Solution {
|
||||
@ -351,6 +378,7 @@ object Solution {
|
||||
// 最后返回栈顶,不需要加return关键字
|
||||
stack.pop()
|
||||
}
|
||||
|
||||
}
|
||||
```
|
||||
-----------------------
|
||||
|
@ -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>
|
||||
|
@ -502,5 +502,35 @@ func combinationSum3(_ count: Int, _ targetSum: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable
|
||||
def combinationSum3(k: Int, n: Int): List[List[Int]] = {
|
||||
var result = mutable.ListBuffer[List[Int]]()
|
||||
var path = mutable.ListBuffer[Int]()
|
||||
|
||||
def backtracking(k: Int, n: Int, sum: Int, startIndex: Int): Unit = {
|
||||
if (sum > n) return // 剪枝,如果sum>目标和,就返回
|
||||
if (sum == n && path.size == k) {
|
||||
result.append(path.toList)
|
||||
return
|
||||
}
|
||||
// 剪枝
|
||||
for (i <- startIndex to (9 - (k - path.size) + 1)) {
|
||||
path.append(i)
|
||||
backtracking(k, n, sum + i, i + 1)
|
||||
path = path.take(path.size - 1)
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(k, n, 0, 1) // 调用递归方法
|
||||
result.toList // 最终返回结果集的List形式
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -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;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
-----------------------
|
||||
|
@ -381,7 +381,36 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
|
||||
};
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
递归:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
|
||||
// scala中每个关键字都有其返回值,于是可以不写return
|
||||
if (root.value > p.value && root.value > q.value) lowestCommonAncestor(root.left, p, q)
|
||||
else if (root.value < p.value && root.value < q.value) lowestCommonAncestor(root.right, p, q)
|
||||
else root
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
迭代:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
|
||||
var curNode = root // 因为root是不可变量,所以要赋值给curNode一个可变量
|
||||
while(curNode != null){
|
||||
if(curNode.value > p.value && curNode.value > q.value) curNode = curNode.left
|
||||
else if(curNode.value < p.value && curNode.value < q.value) curNode = curNode.right
|
||||
else return curNode
|
||||
}
|
||||
null
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
|
@ -343,7 +343,25 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
|
||||
};
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
|
||||
// 递归结束条件
|
||||
if (root == null || root == p || root == q) {
|
||||
return root
|
||||
}
|
||||
|
||||
var left = lowestCommonAncestor(root.left, p, q)
|
||||
var right = lowestCommonAncestor(root.right, p, q)
|
||||
|
||||
if (left != null && right != null) return root
|
||||
if (left == null) return right
|
||||
left
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -654,8 +654,7 @@ object Solution {
|
||||
// 最终返回res,return关键字可以省略
|
||||
res
|
||||
}
|
||||
|
||||
}
|
||||
}
|
||||
|
||||
class MyQueue {
|
||||
var queue = ArrayBuffer[Int]()
|
||||
@ -678,5 +677,84 @@ class MyQueue {
|
||||
def peek(): Int = queue.head
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
PHP:
|
||||
```php
|
||||
class Solution {
|
||||
/**
|
||||
* @param Integer[] $nums
|
||||
* @param Integer $k
|
||||
* @return Integer[]
|
||||
*/
|
||||
function maxSlidingWindow($nums, $k) {
|
||||
$myQueue = new MyQueue();
|
||||
// 先将前k的元素放进队列
|
||||
for ($i = 0; $i < $k; $i++) {
|
||||
$myQueue->push($nums[$i]);
|
||||
}
|
||||
|
||||
$result = [];
|
||||
$result[] = $myQueue->max(); // result 记录前k的元素的最大值
|
||||
|
||||
for ($i = $k; $i < count($nums); $i++) {
|
||||
$myQueue->pop($nums[$i - $k]); // 滑动窗口移除最前面元素
|
||||
$myQueue->push($nums[$i]); // 滑动窗口前加入最后面的元素
|
||||
$result[]= $myQueue->max(); // 记录对应的最大值
|
||||
}
|
||||
return $result;
|
||||
}
|
||||
|
||||
}
|
||||
|
||||
// 单调对列构建
|
||||
class MyQueue{
|
||||
private $queue;
|
||||
|
||||
public function __construct(){
|
||||
$this->queue = new SplQueue(); //底层是双向链表实现。
|
||||
}
|
||||
|
||||
public function pop($v){
|
||||
// 判断当前对列是否为空
|
||||
// 比较当前要弹出的数值是否等于队列出口元素的数值,如果相等则弹出。
|
||||
// bottom 从链表前端查看元素, dequeue 从双向链表的开头移动一个节点
|
||||
if(!$this->queue->isEmpty() && $v == $this->queue->bottom()){
|
||||
$this->queue->dequeue(); //弹出队列
|
||||
}
|
||||
}
|
||||
|
||||
public function push($v){
|
||||
// 判断当前对列是否为空
|
||||
// 如果push的数值大于入口元素的数值,那么就将队列后端的数值弹出,直到push的数值小于等于队列入口元素的数值为止。
|
||||
// 这样就保持了队列里的数值是单调从大到小的了。
|
||||
while (!$this->queue->isEmpty() && $v > $this->queue->top()) {
|
||||
$this->queue->pop(); // pop从链表末尾弹出一个元素,
|
||||
}
|
||||
$this->queue->enqueue($v);
|
||||
}
|
||||
|
||||
// 查询当前队列里的最大值 直接返回队首
|
||||
public function max(){
|
||||
// bottom 从链表前端查看元素, top从链表末尾查看元素
|
||||
return $this->queue->bottom();
|
||||
}
|
||||
|
||||
// 辅助理解: 打印队列元素
|
||||
public function println(){
|
||||
// "迭代器移动到链表头部": 可理解为从头遍历链表元素做准备。
|
||||
// 【PHP中没有指针概念,所以就没说指针。从数据结构上理解,就是把指针指向链表头部】
|
||||
$this->queue->rewind();
|
||||
|
||||
echo "Println: ";
|
||||
while($this->queue->valid()){
|
||||
echo $this->queue->current()," -> ";
|
||||
$this->queue->next();
|
||||
}
|
||||
echo "\n";
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -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;
|
||||
};
|
||||
```
|
||||
|
||||
|
@ -582,7 +582,35 @@ function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
|
||||
};
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def deleteNode(root: TreeNode, key: Int): TreeNode = {
|
||||
if (root == null) return root // 第一种情况,没找到删除的节点,遍历到空节点直接返回
|
||||
if (root.value == key) {
|
||||
// 第二种情况: 左右孩子都为空,直接删除节点,返回null
|
||||
if (root.left == null && root.right == null) return null
|
||||
// 第三种情况: 左孩子为空,右孩子不为空,右孩子补位
|
||||
else if (root.left == null && root.right != null) return root.right
|
||||
// 第四种情况: 左孩子不为空,右孩子为空,左孩子补位
|
||||
else if (root.left != null && root.right == null) return root.left
|
||||
// 第五种情况: 左右孩子都不为空,将删除节点的左子树头节点(左孩子)放到
|
||||
// 右子树的最左边节点的左孩子上,返回删除节点的右孩子
|
||||
else {
|
||||
var tmp = root.right
|
||||
while (tmp.left != null) tmp = tmp.left
|
||||
tmp.left = root.left
|
||||
return root.right
|
||||
}
|
||||
}
|
||||
if (root.value > key) root.left = deleteNode(root.left, key)
|
||||
if (root.value < key) root.right = deleteNode(root.right, key)
|
||||
|
||||
root // 返回根节点,return关键字可以省略
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -352,6 +352,24 @@ function convertBST(root: TreeNode | null): TreeNode | null {
|
||||
};
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def convertBST(root: TreeNode): TreeNode = {
|
||||
var sum = 0
|
||||
def convert(node: TreeNode): Unit = {
|
||||
if (node == null) return
|
||||
convert(node.right)
|
||||
sum += node.value
|
||||
node.value = sum
|
||||
convert(node.left)
|
||||
}
|
||||
convert(root)
|
||||
root
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
|
@ -453,7 +453,21 @@ function trimBST(root: TreeNode | null, low: number, high: number): TreeNode | n
|
||||
};
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
递归法:
|
||||
```scala
|
||||
object Solution {
|
||||
def trimBST(root: TreeNode, low: Int, high: Int): TreeNode = {
|
||||
if (root == null) return null
|
||||
if (root.value < low) return trimBST(root.right, low, high)
|
||||
if (root.value > high) return trimBST(root.left, low, high)
|
||||
root.left = trimBST(root.left, low, high)
|
||||
root.right = trimBST(root.right, low, high)
|
||||
root
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
|
@ -585,6 +585,43 @@ function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
|
||||
```
|
||||
|
||||
|
||||
## Scala
|
||||
|
||||
递归:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
|
||||
if (root == null) return new TreeNode(`val`)
|
||||
if (`val` < root.value) root.left = insertIntoBST(root.left, `val`)
|
||||
else root.right = insertIntoBST(root.right, `val`)
|
||||
root // 返回根节点
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
迭代:
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
|
||||
if (root == null) {
|
||||
return new TreeNode(`val`)
|
||||
}
|
||||
var parent = root // 记录当前节点的父节点
|
||||
var curNode = root
|
||||
while (curNode != null) {
|
||||
parent = curNode
|
||||
if(`val` < curNode.value) curNode = curNode.left
|
||||
else curNode = curNode.right
|
||||
}
|
||||
if(`val` < parent.value) parent.left = new TreeNode(`val`)
|
||||
else parent.right = new TreeNode(`val`)
|
||||
root // 最终返回根节点
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -613,6 +613,36 @@ public class Solution{
|
||||
}
|
||||
```
|
||||
|
||||
**Kotlin:**
|
||||
```kotlin
|
||||
class Solution {
|
||||
fun search(nums: IntArray, target: Int): Int {
|
||||
// leftBorder
|
||||
var left:Int = 0
|
||||
// rightBorder
|
||||
var right:Int = nums.size - 1
|
||||
// 使用左闭右闭区间
|
||||
while (left <= right) {
|
||||
var middle:Int = left + (right - left)/2
|
||||
// taget 在左边
|
||||
if (nums[middle] > target) {
|
||||
right = middle - 1
|
||||
}
|
||||
else {
|
||||
// target 在右边
|
||||
if (nums[middle] < target) {
|
||||
left = middle + 1
|
||||
}
|
||||
// 找到了,返回
|
||||
else return middle
|
||||
}
|
||||
}
|
||||
// 没找到,返回
|
||||
return -1
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
**Kotlin:**
|
||||
|
@ -104,8 +104,9 @@ public:
|
||||
// 在第index个节点之前插入一个新节点,例如index为0,那么新插入的节点为链表的新头节点。
|
||||
// 如果index 等于链表的长度,则说明是新插入的节点为链表的尾结点
|
||||
// 如果index大于链表的长度,则返回空
|
||||
// 如果index小于0,则置为0,作为链表的新头节点。
|
||||
void addAtIndex(int index, int val) {
|
||||
if (index > _size) {
|
||||
if (index > _size || index < 0) {
|
||||
return;
|
||||
}
|
||||
LinkedNode* newNode = new LinkedNode(val);
|
||||
|
@ -260,6 +260,75 @@ var sortArrayByParityII = function(nums) {
|
||||
};
|
||||
```
|
||||
|
||||
### TypeScript
|
||||
|
||||
> 方法一:
|
||||
|
||||
```typescript
|
||||
function sortArrayByParityII(nums: number[]): number[] {
|
||||
const evenArr: number[] = [],
|
||||
oddArr: number[] = [];
|
||||
for (let num of nums) {
|
||||
if (num % 2 === 0) {
|
||||
evenArr.push(num);
|
||||
} else {
|
||||
oddArr.push(num);
|
||||
}
|
||||
}
|
||||
const resArr: number[] = [];
|
||||
for (let i = 0, length = nums.length / 2; i < length; i++) {
|
||||
resArr.push(evenArr[i]);
|
||||
resArr.push(oddArr[i]);
|
||||
}
|
||||
return resArr;
|
||||
};
|
||||
```
|
||||
|
||||
> 方法二:
|
||||
|
||||
```typescript
|
||||
function sortArrayByParityII(nums: number[]): number[] {
|
||||
const length: number = nums.length;
|
||||
const resArr: number[] = [];
|
||||
let evenIndex: number = 0,
|
||||
oddIndex: number = 1;
|
||||
for (let i = 0; i < length; i++) {
|
||||
if (nums[i] % 2 === 0) {
|
||||
resArr[evenIndex] = nums[i];
|
||||
evenIndex += 2;
|
||||
} else {
|
||||
resArr[oddIndex] = nums[i];
|
||||
oddIndex += 2;
|
||||
}
|
||||
}
|
||||
return resArr;
|
||||
};
|
||||
```
|
||||
|
||||
> 方法三:
|
||||
|
||||
```typescript
|
||||
function sortArrayByParityII(nums: number[]): number[] {
|
||||
const length: number = nums.length;
|
||||
let oddIndex: number = 1;
|
||||
for (let evenIndex = 0; evenIndex < length; evenIndex += 2) {
|
||||
if (nums[evenIndex] % 2 === 1) {
|
||||
// 在偶数位遇到了奇数
|
||||
while (oddIndex < length && nums[oddIndex] % 2 === 1) {
|
||||
oddIndex += 2;
|
||||
}
|
||||
// 在奇数位遇到了偶数,交换
|
||||
let temp = nums[evenIndex];
|
||||
nums[evenIndex] = nums[oddIndex];
|
||||
nums[oddIndex] = temp;
|
||||
}
|
||||
}
|
||||
return nums;
|
||||
};
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -421,6 +421,24 @@ object Solution {
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
C#:
|
||||
```csharp
|
||||
public class Solution {
|
||||
public int[] SortedSquares(int[] nums) {
|
||||
int k = nums.Length - 1;
|
||||
int[] result = new int[nums.Length];
|
||||
for (int i = 0, j = nums.Length - 1;i <= j;){
|
||||
if (nums[i] * nums[i] < nums[j] * nums[j]) {
|
||||
result[k--] = nums[j] * nums[j];
|
||||
j--;
|
||||
} else {
|
||||
result[k--] = nums[i] * nums[i];
|
||||
i++;
|
||||
}
|
||||
}
|
||||
return result;
|
||||
}
|
||||
}
|
||||
```
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -356,9 +356,13 @@ func test_2_wei_bag_problem1(weight, value []int, bagweight int) int {
|
||||
// 递推公式
|
||||
for i := 1; i < len(weight); i++ {
|
||||
//正序,也可以倒序
|
||||
for j := weight[i];j<= bagweight ; j++ {
|
||||
dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
|
||||
}
|
||||
for j := 0; j <= bagweight; j++ {
|
||||
if j < weight[i] {
|
||||
dp[i][j] = dp[i-1][j]
|
||||
} else {
|
||||
dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
|
||||
}
|
||||
}
|
||||
}
|
||||
return dp[len(weight)-1][bagweight]
|
||||
}
|
||||
|
Reference in New Issue
Block a user