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Merge branch 'master' of github.com:youngyangyang04/leetcode-master

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74
problems/0015.三数之和.md
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@ -435,5 +435,79 @@ func threeSum(_ nums: [Int]) -> [[Int]] {
}
```
C:
```C
//qsort辅助cmp函数
int cmp(const void* ptr1, const void* ptr2) {
return *((int*)ptr1) > *((int*)ptr2);
}
int** threeSum(int* nums, int numsSize, int* returnSize, int** returnColumnSizes) {
//开辟ans数组空间
int **ans = (int**)malloc(sizeof(int*) * 18000);
int ansTop = 0;
//若传入nums数组大小小于3，则需要返回数组大小为0
if(numsSize < 3) {
*returnSize = 0;
return ans;
}
//对nums数组进行排序
qsort(nums, numsSize, sizeof(int), cmp);
int i;
//用for循环遍历数组，结束条件为i < numsSize - 2(因为要预留左右指针的位置)
for(i = 0; i < numsSize - 2; i++) {
//若当前i指向元素>0，则代表left和right以及i的和大于0。直接break
if(nums[i] > 0)
break;
//去重：i > 0 && nums[i] == nums[i-1]
if(i > 0 && nums[i] == nums[i-1])
continue;
//定义左指针和右指针
int left = i + 1;
int right = numsSize - 1;
//当右指针比左指针大时进行循环
while(right > left) {
//求出三数之和
int sum = nums[right] + nums[left] + nums[i];
//若和小于0，则左指针+1（因为左指针右边的数比当前所指元素大）
if(sum < 0)
left++;
//若和大于0，则将右指针-1
else if(sum > 0)
right--;
//若和等于0
else {
//开辟一个大小为3的数组空间，存入nums[i], nums[left]和nums[right]
int* arr = (int*)malloc(sizeof(int) * 3);
arr[0] = nums[i];
arr[1] = nums[left];
arr[2] = nums[right];
//将开辟数组存入ans中
ans[ansTop++] = arr;
//去重
while(right > left && nums[right] == nums[right - 1])
right--;
while(left < right && nums[left] == nums[left + 1])
left++;
//更新左右指针
left++;
right--;
}
}
}
//设定返回的数组大小
*returnSize = ansTop;
*returnColumnSizes = (int*)malloc(sizeof(int) * ansTop);
int z;
for(z = 0; z < ansTop; z++) {
(*returnColumnSizes)[z] = 3;
}
return ans;
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

3
problems/0019.删除链表的倒数第N个节点.md
View File

@ -173,8 +173,7 @@ var removeNthFromEnd = function(head, n) {
let ret = new ListNode(0, head),
slow = fast = ret;
while(n--) fast = fast.next;
if(!fast) return ret.next;
while (fast.next) {
while (fast.next !== null) {
fast = fast.next;
slow = slow.next
};

31
problems/0040.组合总和II.md
View File

@ -457,6 +457,37 @@ var combinationSum2 = function(candidates, target) {
}
};
```
**使用used去重**
```js
var combinationSum2 = function(candidates, target) {
let res = [];
let path = [];
let total = 0;
const len = candidates.length;
candidates.sort((a, b) => a - b);
let used = new Array(len).fill(false);
const backtracking = (startIndex) => {
if (total === target) {
res.push([...path]);
return;
}
for(let i = startIndex; i < len && total < target; i++) {
const cur = candidates[i];
if (cur > target - total || (i > 0 && cur === candidates[i - 1] && !used[i - 1])) continue;
path.push(cur);
total += cur;
used[i] = true;
backtracking(i + 1);
path.pop();
total -= cur;
used[i] = false;
}
}
backtracking(0);
return res;
};
```
## C
```c

29
problems/0098.验证二叉搜索树.md
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@ -414,19 +414,24 @@ class Solution:
import "math"
func isValidBST(root *TreeNode) bool {
if root == nil {
return true
}
return isBST(root, math.MinInt64, math.MaxFloat64)
// 二叉搜索树也可以是空树
if root == nil {
return true
}
// 由题目中的数据限制可以得出min和max
return check(root,math.MinInt64,math.MaxInt64)
}
func isBST(root *TreeNode, min, max int) bool {
if root == nil {
return true
}
if min >= root.Val || max <= root.Val {
return false
}
return isBST(root.Left, min, root.Val) && isBST(root.Right, root.Val, max)
func check(node *TreeNode,min,max int64) bool {
if node == nil {
return true
}
if min >= int64(node.Val) || max <= int64(node.Val) {
return false
}
// 分别对左子树和右子树递归判断，如果左子树和右子树都符合则返回true
return check(node.Right,int64(node.Val),max) && check(node.Left,min,int64(node.Val))
}
```
```go

32
problems/0150.逆波兰表达式求值.md
View File

@ -241,6 +241,38 @@ class Solution:
```
Swift：
```Swift
func evalRPN(_ tokens: [String]) -> Int {
var stack = [Int]()
for c in tokens {
let v = Int(c)
if let num = v {
// 遇到数字直接入栈
stack.append(num)
} else {
// 遇到运算符, 取出栈顶两元素计算, 结果压栈
var res: Int = 0
let num2 = stack.popLast()!
let num1 = stack.popLast()!
switch c {
case "+":
res = num1 + num2
case "-":
res = num1 - num2
case "*":
res = num1 * num2
case "/":
res = num1 / num2
default:
break
}
stack.append(res)
}
}
return stack.last!
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

18
problems/0205.同构字符串.md
View File

@ -90,6 +90,24 @@ class Solution {
## Python
```python
class Solution:
def isIsomorphic(self, s: str, t: str) -> bool:
default_dict1 = defaultdict(str)
default_dict2 = defaultdict(str)
if len(s) != len(t): return false
for i in range(len(s)):
if not default_dict1[s[i]]:
default_dict1[s[i]] = t[i]
if not default_dict2[t[i]]:
default_dict2[t[i]] = s[i]
if default_dict1[s[i]] != t[i] or default_dict2[t[i]] != s[i]:
return False
return True
```
## Go

40
problems/0236.二叉树的最近公共祖先.md
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@ -221,42 +221,30 @@ public:
## Java
```Java
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
return lowestCommonAncestor1(root, p, q);
}
public TreeNode lowestCommonAncestor1(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q) {
if (root == null || root == p || root == q) { // 递归结束条件
return root;
}
TreeNode left = lowestCommonAncestor1(root.left, p, q);
TreeNode right = lowestCommonAncestor1(root.right, p, q);
if (left != null && right != null) {// 左右子树分别找到了，说明此时的root就是要求的结果
return root;
}
if (left == null) {
// 后序遍历
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if(left == null && right == null) { // 若未找到节点 p 或 q
return null;
}else if(left == null && right != null) { // 若找到一个节点
return right;
}else if(left != null && right == null) { // 若找到一个节点
return left;
}else { // 若找到两个节点
return root;
}
return left;
}
}
```
```java
// 代码精简版
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root.val == p.val ||root.val == q.val) return root;
TreeNode left = lowestCommonAncestor(root.left,p,q);
TreeNode right = lowestCommonAncestor(root.right,p,q);
if (left != null && right != null) return root;
else if (left == null && right != null) return right;
else if (left != null && right == null) return left;
else return null;
}
}
```
## Python

106
problems/0239.滑动窗口最大值.md
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@ -418,6 +418,112 @@ var maxSlidingWindow = function (nums, k) {
};
```
Swift:
```Swift
/// 双向链表
class DoublyListNode {
var head: DoublyListNode?
var tail: DoublyListNode?
var next: DoublyListNode?
var pre: DoublyListNode?
var value: Int = 0
init(_ value: Int = 0) {
self.value = value
}
func isEmpty() -> Bool {
return self.head == nil
}
func first() -> Int? {
return self.head?.value
}
func last() -> Int? {
return self.tail?.value
}
func removeFirst() {
if isEmpty() {
return
}
let next = self.head!.next
self.head?.next = nil// 移除首节点
next?.pre = nil
self.head = next
}
func removeLast() {
if let tail = self.tail {
if let pre = tail.pre {
self.tail?.pre = nil
pre.next = nil
self.tail = pre
} else {
self.head = nil
self.tail = nil
}
}
}
func append(_ value: Int) {
let node = DoublyListNode(value)
if self.head != nil {
node.pre = self.tail
self.tail?.next = node
self.tail = node
} else {
self.head = node
self.tail = node
self.pre = nil
self.next = nil
}
}
}
// 单调队列, 从大到小
class MyQueue {
// var queue: [Int]!// 用数组会超时
var queue: DoublyListNode!
init() {
// queue = [Int]()
queue = DoublyListNode()
}
// 滑动窗口时弹出第一个元素, 如果相等再弹出
func pop(x: Int) {
if !queue.isEmpty() && front() == x {
queue.removeFirst()
}
}
// 滑动窗口时添加下一个元素, 移除队尾比 x 小的元素 始终保证队头 > 队尾
func push(x: Int) {
while !queue.isEmpty() && queue.last()! < x {
queue.removeLast()
}
queue.append(x)
}
// 此时队头就是滑动窗口最大值
func front() -> Int {
return queue.first() ?? -1
}
}
class Solution {
func maxSlidingWindow(_ nums: [Int], _ k: Int) -> [Int] {
// 存放结果
var res = [Int]()
let queue = MyQueue()
// 先将前K个元素放入队列
for i in 0 ..< k {
queue.push(x: nums[i])
}
// 添加当前队列最大值到结果数组
res.append(queue.front())
for i in k ..< nums.count {
// 滑动窗口移除最前面元素
queue.pop(x: nums[i - k])
// 滑动窗口添加下一个元素
queue.push(x: nums[i])
// 保存当前队列最大值
res.append(queue.front())
}
return res
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

2
problems/0416.分割等和子集.md
View File

@ -68,7 +68,7 @@
* 背包的体积为sum / 2
* 背包要放入的商品（集合里的元素）重量为 元素的数值，价值也为元素的数值
* 背包如何正好装满，说明找到了总和为 sum / 2 的子集。
* 背包如果正好装满，说明找到了总和为 sum / 2 的子集。
* 背包中每一个元素是不可重复放入。
以上分析完，我们就可以套用01背包，来解决这个问题了。

9
problems/0701.二叉搜索树中的插入操作.md
View File

@ -236,16 +236,13 @@ class Solution {
```java
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
return buildTree(root, val);
}
public TreeNode buildTree(TreeNode root, int val){
if (root == null) // 如果当前节点为空，也就意味着val找到了合适的位置，此时创建节点直接返回。
return new TreeNode(val);
if (root.val < val){
root.right = buildTree(root.right, val); // 递归创建右子树
root.right = insertIntoBST(root.right, val); // 递归创建右子树
}else if (root.val > val){
root.left = buildTree(root.left, val); // 递归创建左子树
root.left = insertIntoBST(root.left, val); // 递归创建左子树
}
return root;
}

10
problems/0704.二分查找.md
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@ -276,11 +276,9 @@ func search(nums []int, target int) int {
```
**JavaScript:**
（版本一）左闭右闭区间
```js
// （版本一）左闭右闭区间
/**
* @param {number[]} nums
* @param {number} target
@ -302,9 +300,10 @@ var search = function(nums, target) {
}
return -1;
};
```
（版本二）左闭右开区间
// （版本二）左闭右开区间
```js
/**
* @param {number[]} nums
* @param {number} target
@ -325,7 +324,6 @@ var search = function(nums, target) {
}
return -1;
};
```
**Ruby:**

37
problems/0977.有序数组的平方.md
View File

@ -199,27 +199,26 @@ impl Solution {
Javascript：
```Javascript
/**
* @desc two pointers solution
* @link https://leetcode-cn.com/problems/squares-of-a-sorted-array/
* @param nums Array e.g. [-4,-1,0,3,10]
* @return {array} e.g. [0,1,9,16,100]
* @param {number[]} nums
* @return {number[]}
*/
const sortedSquares = function (nums) {
let res = []
for (let i = 0, j = nums.length - 1; i <= j;) {
const left = Math.abs(nums[i])
const right = Math.abs(nums[j])
if (right > left) {
// push element to the front of the array
res.unshift(right * right)
j--
} else {
res.unshift(left * left)
i++
}
var sortedSquares = function(nums) {
let n = nums.length;
let res = new Array(n).fill(0);
let i = 0, j = n - 1, k = n - 1;
while (i <= j) {
let left = nums[i] * nums[i],
right = nums[j] * nums[j];
if (left < right) {
res[k--] = right;
j--;
} else {
res[k--] = left;
i++;
}
}
return res
}
return res;
};
```
Swift:

2
problems/算法模板.md
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@ -460,7 +460,7 @@ var levelOrder = function (root) {
for (let i = 0; i < n; i++) {
let node = queue.shift();
temp.push(node.val);
node.left &&queue.push(node.left);
node.left && queue.push(node.left);
node.right && queue.push(node.right);
}
res.push(temp);