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Merge branch 'master' into master

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ironartisan
2021-09-04 20:00:43 +08:00
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25
problems/0001.两数之和.md
View File

@ -224,6 +224,31 @@ func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
}
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums
* @param Integer $target
* @return Integer[]
*/
function twoSum($nums, $target) {
if (count($nums) == 0) {
return [];
}
$table = [];
for ($i = 0; $i < count($nums); $i++) {
$temp = $target - $nums[$i];
if (isset($table[$temp])) {
return [$table[$temp], $i];
}
$table[$nums[$i]] = $i;
}
return [];
}
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频：[代码随想录](https://space.bilibili.com/525438321)

40
problems/0015.三数之和.md
View File

@ -393,6 +393,46 @@ function threeSum(array $nums): array
}
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums
* @return Integer[][]
*/
function threeSum($nums) {
$res = [];
sort($nums);
for ($i = 0; $i < count($nums); $i++) {
if ($nums[$i] > 0) {
return $res;
}
if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
continue;
}
$left = $i + 1;
$right = count($nums) - 1;
while ($left < $right) {
$sum = $nums[$i] + $nums[$left] + $nums[$right];
if ($sum < 0) {
$left++;
}
else if ($sum > 0) {
$right--;
}
else {
$res[] = [$nums[$i], $nums[$left], $nums[$right]];
while ($left < $right && $nums[$left] == $nums[$left + 1]) $left++;
while ($left < $right && $nums[$right] == $nums[$right - 1]) $right--;
$left++;
$right--;
}
}
}
return $res;
}
}
```
-----------------------

43
problems/0018.四数之和.md
View File

@ -310,6 +310,49 @@ var fourSum = function(nums, target) {
};
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums
* @param Integer $target
* @return Integer[][]
*/
function fourSum($nums, $target) {
$res = [];
sort($nums);
for ($i = 0; $i < count($nums); $i++) {
if ($i > 0 && $nums[$i] == $nums[$i - 1]) {
continue;
}
for ($j = $i + 1; $j < count($nums); $j++) {
if ($j > $i + 1 && $nums[$j] == $nums[$j - 1]) {
continue;
}
$left = $j + 1;
$right = count($nums) - 1;
while ($left < $right) {
$sum = $nums[$i] + $nums[$j] + $nums[$left] + $nums[$right];
if ($sum < $target) {
$left++;
}
else if ($sum > $target) {
$right--;
}
else {
$res[] = [$nums[$i], $nums[$j], $nums[$left], $nums[$right]];
while ($left < $right && $nums[$left] == $nums[$left+1]) $left++;
while ($left < $right && $nums[$right] == $nums[$right-1]) $right--;
$left++;
$right--;
}
}
}
}
return $res;
}
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

51
problems/0077.组合优化.md
View File

@ -242,8 +242,59 @@ var combine = function(n, k) {
};
```
C:
```c
int* path;
int pathTop;
int** ans;
int ansTop;
void backtracking(int n, int k,int startIndex) {
//当path中元素个数为k个时，我们需要将path数组放入ans二维数组中
if(pathTop == k) {
//path数组为我们动态申请，若直接将其地址放入二维数组，path数组中的值会随着我们回溯而逐渐变化
//因此创建新的数组存储path中的值
int* temp = (int*)malloc(sizeof(int) * k);
int i;
for(i = 0; i < k; i++) {
temp[i] = path[i];
}
ans[ansTop++] = temp;
return ;
}
int j;
for(j = startIndex; j <= n- (k - pathTop) + 1;j++) {
//将当前结点放入path数组
path[pathTop++] = j;
//进行递归
backtracking(n, k, j + 1);
//进行回溯，将数组最上层结点弹出
pathTop--;
}
}
int** combine(int n, int k, int* returnSize, int** returnColumnSizes){
//path数组存储符合条件的结果
path = (int*)malloc(sizeof(int) * k);
//ans二维数组存储符合条件的结果数组的集合。（数组足够大，避免极端情况）
ans = (int**)malloc(sizeof(int*) * 10000);
pathTop = ansTop = 0;
//回溯算法
backtracking(n, k, 1);
//最后的返回大小为ans数组大小
*returnSize = ansTop;
//returnColumnSizes数组存储ans二维数组对应下标中一维数组的长度（都为k）
*returnColumnSizes = (int*)malloc(sizeof(int) *(*returnSize));
int i;
for(i = 0; i < *returnSize; i++) {
(*returnColumnSizes)[i] = k;
}
//返回ans二维数组
return ans;
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

179
problems/0102.二叉树的层序遍历.md
View File

@ -87,27 +87,33 @@ public:
python代码：
```python
# 迭代法
```python3
class Solution:
"""二叉树层序遍历迭代解法"""
def levelOrder(self, root: TreeNode) -> List[List[int]]:
results = []
if not root:
return []
return results
from collections import deque
que = deque([root])
while que:
size = len(que)
result = []
for _ in range(size):
cur = que.popleft()
result.append(cur.val)
if cur.left:
que.append(cur.left)
if cur.right:
que.append(cur.right)
results.append(result)
queue = [root]
out_list = []
while queue:
length = len(queue)
in_list = []
for _ in range(length):
curnode = queue.pop(0) # （默认移除列表最后一个元素）这里需要移除队列最头上的那个
in_list.append(curnode.val)
if curnode.left: queue.append(curnode.left)
if curnode.right: queue.append(curnode.right)
out_list.append(in_list)
return out_list
return results
```
```python
# 递归法
@ -288,29 +294,29 @@ python代码：
```python
class Solution:
"""二叉树层序遍历II迭代解法"""
def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
results = []
if not root:
return []
quene = [root]
out_list = []
return results
while quene:
in_list = []
for _ in range(len(quene)):
node = quene.pop(0)
in_list.append(node.val)
if node.left:
quene.append(node.left)
if node.right:
quene.append(node.right)
out_list.append(in_list)
out_list.reverse()
return out_list
# 执行用时：36 ms, 在所有 Python3 提交中击败了92.00%的用户
# 内存消耗：15.2 MB, 在所有 Python3 提交中击败了63.76%的用户
from collections import deque
que = deque([root])
while que:
result = []
for _ in range(len(que)):
cur = que.popleft()
result.append(cur.val)
if cur.left:
que.append(cur.left)
if cur.right:
que.append(cur.right)
results.append(result)
results.reverse()
return results
```
Java：
@ -642,32 +648,29 @@ python代码：
```python
class Solution:
"""二叉树层平均值迭代解法"""
def averageOfLevels(self, root: TreeNode) -> List[float]:
results = []
if not root:
return []
return results
quene = deque([root])
out_list = []
while quene:
in_list = []
for _ in range(len(quene)):
node = quene.popleft()
in_list.append(node.val)
if node.left:
quene.append(node.left)
if node.right:
quene.append(node.right)
out_list.append(in_list)
out_list = map(lambda x: sum(x) / len(x), out_list)
return out_list
from collections import deque
que = deque([root])
# 执行用时：56 ms, 在所有 Python3 提交中击败了81.48%的用户
# 内存消耗：17 MB, 在所有 Python3 提交中击败了89.68%的用户
while que:
size = len(que)
sum_ = 0
for _ in range(size):
cur = que.popleft()
sum_ += cur.val
if cur.left:
que.append(cur.left)
if cur.right:
que.append(cur.right)
results.append(sum_ / size)
return results
```
java:
@ -837,52 +840,28 @@ public:
python代码：
```python
class Solution:
"""N叉树的层序遍历迭代法"""
def levelOrder(self, root: 'Node') -> List[List[int]]:
results = []
if not root:
return []
return results
quene = deque([root])
out_list = []
while quene:
in_list = []
for _ in range(len(quene)):
node = quene.popleft()
in_list.append(node.val)
if node.children:
# 这个地方要用extend而不是append，我们看下面的例子：
# In [18]: alist=[]
# In [19]: alist.append([1,2,3])
# In [20]: alist
# Out[20]: [[1, 2, 3]]
# In [21]: alist.extend([4,5,6])
# In [22]: alist
# Out[22]: [[1, 2, 3], 4, 5, 6]
# 可以看到extend对要添加的list进行了一个解包操作
# print(root.children)，可以得到children是一个包含
# 孩子节点地址的list，我们使用for遍历quene的时候，
# 希望quene是一个单层list，所以要用extend
# 使用extend的情况，如果print(quene),结果是
# deque([<__main__.Node object at 0x7f60763ae0a0>])
# deque([<__main__.Node object at 0x7f607636e6d0>, <__main__.Node object at 0x7f607636e130>, <__main__.Node object at 0x7f607636e310>])
# deque([<__main__.Node object at 0x7f607636e880>, <__main__.Node object at 0x7f607636ef10>])
# 可以看到是单层list
# 如果使用append，print(quene)的结果是
# deque([<__main__.Node object at 0x7f18907530a0>])
# deque([[<__main__.Node object at 0x7f18907136d0>, <__main__.Node object at 0x7f1890713130>, <__main__.Node object at 0x7f1890713310>]])
# 可以看到是两层list，这样for的遍历就会报错
quene.extend(node.children)
out_list.append(in_list)
from collections import deque
que = deque([root])
return out_list
# 执行用时：60 ms, 在所有 Python3 提交中击败了76.99%的用户
# 内存消耗：16.5 MB, 在所有 Python3 提交中击败了89.19%的用户
while que:
result = []
for _ in range(len(que)):
cur = que.popleft()
result.append(cur.val)
# cur.children 是 Node 对象组成的列表，也可能为 None
if cur.children:
que.extend(cur.children)
results.append(result)
return results
```
java:

31
problems/0202.快乐数.md
View File

@ -223,6 +223,37 @@ func isHappy(_ n: Int) -> Bool {
}
```
PHP:
```php
class Solution {
/**
* @param Integer $n
* @return Boolean
*/
function isHappy($n) {
// use a set to record sum
// whenever there is a duplicated, stop
// == 1 return true, else false
$table = [];
while ($n != 1 && !isset($table[$n])) {
$table[$n] = 1;
$n = self::getNextN($n);
}
return $n == 1;
}
function getNextN(int $n) {
$res = 0;
while ($n > 0) {
$temp = $n % 10;
$res += $temp * $temp;
$n = floor($n / 10);
}
return $res;
}
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频：[代码随想录](https://space.bilibili.com/525438321)

28
problems/0203.移除链表元素.md
View File

@ -331,8 +331,32 @@ func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
}
```
PHP:
```php
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
// 虚拟头+双指针
func removeElements(head *ListNode, val int) *ListNode {
dummyHead := &ListNode{}
dummyHead.Next = head
pred := dummyHead
cur := head
for cur != nil {
if cur.Val == val {
pred.Next = cur.Next
} else {
pred = cur
}
cur = cur.Next
}
return dummyHead.Next
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

35
problems/0242.有效的字母异位词.md
View File

@ -221,6 +221,41 @@ func isAnagram(_ s: String, _ t: String) -> Bool {
}
```
PHP：
```php
class Solution {
/**
* @param String $s
* @param String $t
* @return Boolean
*/
function isAnagram($s, $t) {
if (strlen($s) != strlen($t)) {
return false;
}
$table = [];
for ($i = 0; $i < strlen($s); $i++) {
if (!isset($table[$s[$i]])) {
$table[$s[$i]] = 1;
} else {
$table[$s[$i]]++;
}
if (!isset($table[$t[$i]])) {
$table[$t[$i]] = -1;
} else {
$table[$t[$i]]--;
}
}
foreach ($table as $record) {
if ($record != 0) {
return false;
}
}
return true;
}
}
```
## 相关题目
* 383.赎金信

29
problems/0349.两个数组的交集.md
View File

@ -209,6 +209,35 @@ func intersection(_ nums1: [Int], _ nums2: [Int]) -> [Int] {
}
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums1
* @param Integer[] $nums2
* @return Integer[]
*/
function intersection($nums1, $nums2) {
if (count($nums1) == 0 || count($nums2) == 0) {
return [];
}
$counts = [];
$res = [];
foreach ($nums1 as $num) {
$counts[$num] = 1;
}
foreach ($nums2 as $num) {
if (isset($counts[$num])) {
$res[] = $num;
}
unset($counts[$num]);
}
return $res;
}
}
```
## 相关题目
* 350.两个数组的交集 II

48
problems/0383.赎金信.md
View File

@ -267,6 +267,54 @@ var canConstruct = function(ransomNote, magazine) {
```
PHP:
```php
class Solution {
/**
* @param String $ransomNote
* @param String $magazine
* @return Boolean
*/
function canConstruct($ransomNote, $magazine) {
if (count($ransomNote) > count($magazine)) {
return false;
}
$map = [];
for ($i = 0; $i < strlen($magazine); $i++) {
$map[$magazine[$i]] = ($map[$magazine[$i]] ?? 0) + 1;
}
for ($i = 0; $i < strlen($ransomNote); $i++) {
if (!isset($map[$ransomNote[$i]]) || --$map[$ransomNote[$i]] < 0) {
return false;
}
}
return true;
}
```
Swift：
```swift
func canConstruct(_ ransomNote: String, _ magazine: String) -> Bool {
var record = Array(repeating: 0, count: 26);
let aUnicodeScalarValue = "a".unicodeScalars.first!.value
for unicodeScalar in magazine.unicodeScalars {
// 通过record 记录 magazine 里各个字符出现的次数
let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
record[idx] += 1
}
for unicodeScalar in ransomNote.unicodeScalars {
// 遍历 ransomNote,在record里对应的字符个数做 -- 操作
let idx: Int = Int(unicodeScalar.value - aUnicodeScalarValue)
record[idx] -= 1
// 如果小于零说明在magazine没有
if record[idx] < 0 {
return false
}
}
return true
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

60
problems/0454.四数相加II.md
View File

@ -221,6 +221,66 @@ var fourSumCount = function(nums1, nums2, nums3, nums4) {
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums1
* @param Integer[] $nums2
* @param Integer[] $nums3
* @param Integer[] $nums4
* @return Integer
*/
function fourSumCount($nums1, $nums2, $nums3, $nums4) {
$map = [];
foreach ($nums1 as $n1) {
foreach ($nums2 as $n2) {
$temp = $n1 + $n2;
$map[$temp] = isset($map[$temp]) ? $map[$temp]+1 : 1;
}
}
$count = 0;
foreach ($nums3 as $n3) {
foreach ($nums4 as $n4) {
$temp = 0 - $n3 - $n4;
if (isset($map[$temp])) {
$count += $map[$temp];
}
}
}
return $count;
}
}
```
Swift：
```swift
func fourSumCount(_ nums1: [Int], _ nums2: [Int], _ nums3: [Int], _ nums4: [Int]) -> Int {
// key:a+b的数值，value:a+b数值出现的次数
var map = [Int: Int]()
// 遍历nums1和nums2数组，统计两个数组元素之和，和出现的次数，放到map中
for i in 0 ..< nums1.count {
for j in 0 ..< nums2.count {
let sum1 = nums1[i] + nums2[j]
map[sum1] = (map[sum1] ?? 0) + 1
}
}
// 统计a+b+c+d = 0 出现的次数
var res = 0
// 在遍历大num3和num4数组，找到如果 0-(c+d) 在map中出现过的话，就把map中key对应的value也就是出现次数统计出来。
for i in 0 ..< nums3.count {
for j in 0 ..< nums4.count {
let sum2 = nums3[i] + nums4[j]
let other = 0 - sum2
if map.keys.contains(other) {
res += map[other]!
}
}
}
return res
}
```
-----------------------

5
problems/0455.分发饼干.md
View File

@ -197,11 +197,10 @@ func findContentChildren(g []int, s []int) int {
return child
}
```
Javascript:
​```Javascript
```
var findContentChildren = function(g, s) {
g = g.sort((a, b) => a - b)
s = s.sort((a, b) => a - b)

43
problems/0530.二叉搜索树的最小绝对差.md
View File

@ -265,7 +265,7 @@ func getMinimumDifference(root *TreeNode) int {
```
## JavaScript
递归 先转换为有序数组
```javascript
/**
* Definition for a binary tree node.
@ -297,6 +297,47 @@ var getMinimumDifference = function (root) {
return diff;
};
```
递归 在递归的过程中更新最小值
```js
var getMinimumDifference = function(root) {
let res = Infinity
let preNode = null
// 中序遍历
const inorder = (node) => {
if(!node) return
inorder(node.left)
// 更新res
if(preNode) res = Math.min(res, node.val - preNode.val)
// 记录前一个节点
preNode = node
inorder(node.right)
}
inorder(root)
return res
}
```
迭代 中序遍历
```js
var getMinimumDifference = function(root) {
let stack = []
let cur = root
let res = Infinity
let pre = null
while(cur || stack.length) {
if(cur) {
stack.push(cur)
cur = cur.left
} else {
cur = stack.pop()
if(pre) res = Math.min(res, cur.val - pre.val)
pre = cur
cur = cur.right
}
}
return res
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

35
problems/0977.有序数组的平方.md
View File

@ -270,6 +270,40 @@ def sorted_squares(nums)
end
```
C:
```c
int* sortedSquares(int* nums, int numsSize, int* returnSize){
//返回的数组大小就是原数组大小
*returnSize = numsSize;
//创建两个指针，right指向数组最后一位元素，left指向数组第一位元素
int right = numsSize - 1;
int left = 0;
//最后要返回的结果数组
int* ans = (int*)malloc(sizeof(int) * numsSize);
int index;
for(index = numsSize - 1; index >= 0; index--) {
//左指针指向元素的平方
int lSquare = nums[left] * nums[left];
//右指针指向元素的平方
int rSquare = nums[right] * nums[right];
//若左指针指向元素平方比右指针指向元素平方大，将左指针指向元素平方放入结果数组。左指针右移一位
if(lSquare > rSquare) {
ans[index] = lSquare;
left++;
}
//若右指针指向元素平方比左指针指向元素平方大，将右指针指向元素平方放入结果数组。右指针左移一位
else {
ans[index] = rSquare;
right--;
}
}
//返回结果数组
return ans;
}
```
PHP:
```php
class Solution {
@ -300,6 +334,7 @@ class Solution {
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频：[代码随想录](https://space.bilibili.com/525438321)

4
problems/二叉树的迭代遍历.md
View File

@ -27,7 +27,7 @@
我们先看一下前序遍历。
前序遍历是中左右，每次先处理的是中间节点，那么先将跟节点放入栈中，然后将右孩子加入栈，再加入左孩子。
前序遍历是中左右，每次先处理的是中间节点，那么先将根节点放入栈中，然后将右孩子加入栈，再加入左孩子。
为什么要先加入 右孩子，再加入左孩子呢？ 因为这样出栈的时候才是中左右的顺序。
@ -140,7 +140,7 @@ public:
# 总结
此时我们用迭代法写出了二叉树的前后中序遍历，大家可以看出前序和中序是完全两种代码风格，并不想递归写法那样代码稍做调整，就可以实现前后中序。
此时我们用迭代法写出了二叉树的前后中序遍历，大家可以看出前序和中序是完全两种代码风格，并不像递归写法那样代码稍做调整，就可以实现前后中序。
**这是因为前序遍历中访问节点（遍历节点）和处理节点（将元素放进result数组中）可以同步处理，但是中序就无法做到同步！**