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

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youngyangyang04
2021-09-05 23:19:55 +08:00
parent b2c6d5078a 2c27288ee7
commit 4d55047e01
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42
problems/0001.两数之和.md
View File

@ -206,6 +206,48 @@ function twoSum(array $nums, int $target): array
}
```
Swift：
```swift
func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
var res = [Int]()
var dict = [Int : Int]()
for i in 0 ..< nums.count {
let other = target - nums[i]
if dict.keys.contains(other) {
res.append(i)
res.append(dict[other]!)
return res
}
dict[nums[i]] = i
}
return res
}
```
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)

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

16
problems/0017.电话号码的字母组合.md
View File

@ -322,20 +322,20 @@ python3：
```py
class Solution:
def letterCombinations(self, digits: str) -> List[str]:
self.s = ""
res = []
s = ""
letterMap = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"]
if len(digits) == 0: return res
def backtrack(digits,index):
if not len(digits): return res
def backtrack(digits,index, s):
if index == len(digits):
return res.append(self.s)
return res.append(s)
digit = int(digits[index]) #将index指向的数字转为int
letters = letterMap[digit] #取数字对应的字符集
for i in range(len(letters)):
self.s += letters[i]
backtrack(digits,index + 1) #递归，注意index+1，一下层要处理下一个数字
self.s = self.s[:-1] #回溯
backtrack(digits,0)
s += letters[i]
backtrack(digits, index+1, s) #递归，注意index+1，一下层要处理下一个数字
s = s[:-1] #回溯
backtrack(digits, 0, s)
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)

40
problems/0027.移除元素.md
View File

@ -246,6 +246,46 @@ func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
}
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums
* @param Integer $val
* @return Integer
*/
function removeElement(&$nums, $val) {
if (count($nums) == 0) {
return 0;
}
// 快慢指针
$slow = 0;
for ($fast = 0; $fast < count($nums); $fast++) {
if ($nums[$fast] != $val) {
$nums[$slow] = $nums[$fast];
$slow++;
}
}
return $slow;
}
```
C:
```c
int removeElement(int* nums, int numsSize, int val){
int slow = 0;
for(int fast = 0; fast < numsSize; fast++) {
//若快指针位置的元素不等于要删除的元素
if(nums[fast] != val) {
//将其挪到慢指针指向的位置，慢指针+1
nums[slow++] = nums[fast];
}
}
//最后慢指针的大小就是新的数组的大小
return slow;
}
```
-----------------------
* 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)

26
problems/0046.全排列.md
View File

@ -183,6 +183,32 @@ class Solution {
}
}
```
```java
// 解法2：通过判断path中是否存在数字，排除已经选择的数字
class Solution {
List<List<Integer>> result = new ArrayList<>();
LinkedList<Integer> path = new LinkedList<>();
public List<List<Integer>> permute(int[] nums) {
if (nums.length == 0) return result;
backtrack(nums, path);
return result;
}
public void backtrack(int[] nums, LinkedList<Integer> path) {
if (path.size() == nums.length) {
result.add(new ArrayList<>(path));
}
for (int i =0; i < nums.length; i++) {
// 如果path中已有，则跳过
if (path.contains(nums[i])) {
continue;
}
path.add(nums[i]);
backtrack(nums, path);
path.removeLast();
}
}
}
```
Python：
```python3

22
problems/0056.合并区间.md
View File

@ -157,6 +157,28 @@ class Solution {
}
}
```
```java
// 版本2
class Solution {
public int[][] merge(int[][] intervals) {
LinkedList<int[]> res = new LinkedList<>();
Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0]));
res.add(intervals[0]);
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][0] <= res.getLast()[1]) {
int start = res.getLast()[0];
int end = Math.max(intervals[i][1], res.getLast()[1]);
res.removeLast();
res.add(new int[]{start, end});
}
else {
res.add(intervals[i]);
}
}
return res.toArray(new int[res.size()][]);
}
}
```
Python：
```python

42
problems/0059.螺旋矩阵II.md
View File

@ -426,6 +426,48 @@ impl Solution {
}
```
PHP:
```php
class Solution {
/**
* @param Integer $n
* @return Integer[][]
*/
function generateMatrix($n) {
// 初始化数组
$res = array_fill(0, $n, array_fill(0, $n, 0));
$mid = $loop = floor($n / 2);
$startX = $startY = 0;
$offset = 1;
$count = 1;
while ($loop > 0) {
$i = $startX;
$j = $startY;
for (; $j < $startY + $n - $offset; $j++) {
$res[$i][$j] = $count++;
}
for (; $i < $startX + $n - $offset; $i++) {
$res[$i][$j] = $count++;
}
for (; $j > $startY; $j--) {
$res[$i][$j] = $count++;
}
for (; $i > $startX; $i--) {
$res[$i][$j] = $count++;
}
$startX += 1;
$startY += 1;
$offset += 2;
$loop--;
}
if ($n % 2 == 1) {
$res[$mid][$mid] = $count;
}
return $res;
}
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

53
problems/0077.组合.md
View File

@ -435,6 +435,59 @@ func backtrack(n,k,start int,track []int){
}
```
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 ;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)

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)

195
problems/0102.二叉树的层序遍历.md
View File

@ -87,28 +87,48 @@ public:
python代码：
```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 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)
return results
```
```python
# 递归法
class Solution:
def levelOrder(self, root: TreeNode) -> List[List[int]]:
if not root:
return []
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
res = []
def helper(root, depth):
if not root: return []
if len(res) == depth: res.append([]) # start the current depth
res[depth].append(root.val) # fulfil the current depth
if root.left: helper(root.left, depth + 1) # process child nodes for the next depth
if root.right: helper(root.right, depth + 1)
helper(root, 0)
return res
```
java:
```Java
@ -274,29 +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：
@ -628,32 +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:
@ -823,52 +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:

24
problems/0115.不同的子序列.md
View File

@ -221,6 +221,30 @@ class SolutionDP2:
```
Go：
```go
func numDistinct(s string, t string) int {
dp:= make([][]int,len(s)+1)
for i:=0;i<len(dp);i++{
dp[i] = make([]int,len(t)+1)
}
// 初始化
for i:=0;i<len(dp);i++{
dp[i][0] = 1
}
// dp[0][j] 为 0，默认值，因此不需要初始化
for i:=1;i<len(dp);i++{
for j:=1;j<len(dp[i]);j++{
if s[i-1] == t[j-1]{
dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
}else{
dp[i][j] = dp[i-1][j]
}
}
}
return dp[len(dp)-1][len(dp[0])-1]
}
```
Javascript:
```javascript

51
problems/0134.加油站.md
View File

@ -200,6 +200,7 @@ public:
Java：
```java
// 解法1
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int sum = 0;
@ -221,7 +222,26 @@ class Solution {
}
}
```
```java
// 解法2
class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int curSum = 0;
int totalSum = 0;
int index = 0;
for (int i = 0; i < gas.length; i++) {
curSum += gas[i] - cost[i];
totalSum += gas[i] - cost[i];
if (curSum < 0) {
index = (i + 1) % gas.length ;
curSum = 0;
}
}
if (totalSum < 0) return -1;
return index;
}
}
```
Python：
```python
class Solution:
@ -283,6 +303,35 @@ var canCompleteCircuit = function(gas, cost) {
};
```
C:
```c
int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize){
int curSum = 0;
int i;
int min = INT_MAX;
//遍历整个数组。计算出每站的用油差。并将其与最小累加量比较
for(i = 0; i < gasSize; i++) {
int diff = gas[i] - cost[i];
curSum += diff;
if(curSum < min)
min = curSum;
}
//若汽油总数为负数，代表无法跑完一环。返回-1
if(curSum < 0)
return -1;
//若min大于等于0，说明每一天加油量比用油量多。因此从0出发即可
if(min >= 0)
return 0;
//若累加最小值为负，则找到一个非零元素（加油量大于出油量）出发。返回坐标
for(i = gasSize - 1; i >= 0; i--) {
min+=(gas[i]-cost[i]);
if(min >= 0)
return i;
}
//逻辑上不会返回这个0
return 0;
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

79
problems/0151.翻转字符串里的单词.md
View File

@ -467,6 +467,85 @@ function reverse(strArr, start, end) {
}
```
Swift:
```swift
func reverseWords(_ s: String) -> String {
var stringArr = removeSpace(s)
reverseString(&stringArr, startIndex: 0, endIndex: stringArr.count - 1)
reverseWord(&stringArr)
return String(stringArr)
}
/// 1、移除多余的空格（前后所有的空格，中间只留一个空格）
func removeSpace(_ s: String) -> [Character] {
let ch = Array(s)
var left = 0
var right = ch.count - 1
// 忽略字符串前面的所有空格
while ch[left] == " " {
left += 1
}
// 忽略字符串后面的所有空格
while ch[right] == " " {
right -= 1
}
// 接下来就是要处理中间的多余空格
var lastArr = Array<Character>()
while left <= right {
// 准备加到新字符串当中的字符
let char = ch[left]
// 新的字符串的最后一个字符；或者原字符串中，准备加到新字符串的那个字符；这两个字符当中，只要有一个不是空格，就可以加到新的字符串当中
if char != " " || lastArr[lastArr.count - 1] != " " {
lastArr.append(char)
}
left += 1
}
return lastArr
}
/// 2、反转整个字符串
func reverseString(_ s: inout [Character], startIndex: Int, endIndex: Int) {
var start = startIndex
var end = endIndex
while start < end {
(s[start], s[end]) = (s[end], s[start])
start += 1
end -= 1
}
}
/// 3、再次将字符串里面的单词反转
func reverseWord(_ s: inout [Character]) {
var start = 0
var end = 0
var entry = false
for i in 0..<s.count {
if !entry {
start = i
entry = true
}
if entry && s[i] == " " && s[i - 1] != " " {
end = i - 1
entry = false
reverseString(&s, startIndex: start, endIndex: end)
}
if entry && (i == s.count - 1) && s[i] != " " {
end = i
entry = false
reverseString(&s, startIndex: start, endIndex: end)
}
}
}
```

61
problems/0202.快乐数.md
View File

@ -191,7 +191,68 @@ var isHappy = function(n) {
};
```
Swift：
```swift
// number 每个位置上的数字的平方和
func getSum(_ number: Int) -> Int {
var sum = 0
var num = number
while num > 0 {
let temp = num % 10
sum += (temp * temp)
num /= 10
}
return sum
}
func isHappy(_ n: Int) -> Bool {
var set = Set<Int>()
var num = n
while true {
let sum = self.getSum(num)
if sum == 1 {
return true
}
// 如果这个sum曾经出现过，说明已经陷入了无限循环了
if set.contains(sum) {
return false
} else {
set.insert(sum)
}
num = sum
}
}
```
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)

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)

28
problems/0209.长度最小的子数组.md
View File

@ -264,6 +264,34 @@ impl Solution {
}
```
PHP:
```php
// 双指针 - 滑动窗口
class Solution {
/**
* @param Integer $target
* @param Integer[] $nums
* @return Integer
*/
function minSubArrayLen($target, $nums) {
if (count($nums) < 1) {
return 0;
}
$sum = 0;
$res = PHP_INT_MAX;
$left = 0;
for ($right = 0; $right < count($nums); $right++) {
$sum += $nums[$right];
while ($sum >= $target) {
$res = min($res, $right - $left + 1);
$sum -= $nums[$left];
$left++;
}
}
return $res == PHP_INT_MAX ? 0 : $res;
}
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

22
problems/0222.完全二叉树的节点个数.md
View File

@ -204,7 +204,27 @@ class Solution {
}
}
```
```java
class Solution {
// 迭代法
public int countNodes(TreeNode root) {
if (root == null) return 0;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int result = 0;
while (!queue.isEmpty()) {
int size = queue.size();
while (size -- > 0) {
TreeNode cur = queue.poll();
result++;
if (cur.left != null) queue.offer(cur.left);
if (cur.right != null) queue.offer(cur.right);
}
}
return result;
}
}
```
```java
class Solution {
/**

65
problems/0225.用队列实现栈.md
View File

@ -359,6 +359,71 @@ class MyStack:
Go：
```go
type MyStack struct {
queue []int//创建一个队列
}
/** Initialize your data structure here. */
func Constructor() MyStack {
return MyStack{ //初始化
queue:make([]int,0),
}
}
/** Push element x onto stack. */
func (this *MyStack) Push(x int) {
//添加元素
this.queue=append(this.queue,x)
}
/** Removes the element on top of the stack and returns that element. */
func (this *MyStack) Pop() int {
n:=len(this.queue)-1//判断长度
for n!=0{ //除了最后一个，其余的都重新添加到队列里
val:=this.queue[0]
this.queue=this.queue[1:]
this.queue=append(this.queue,val)
n--
}
//弹出元素
val:=this.queue[0]
this.queue=this.queue[1:]
return val
}
/** Get the top element. */
func (this *MyStack) Top() int {
//利用Pop函数，弹出来的元素重新添加
val:=this.Pop()
this.queue=append(this.queue,val)
return val
}
/** Returns whether the stack is empty. */
func (this *MyStack) Empty() bool {
return len(this.queue)==0
}
/**
* Your MyStack object will be instantiated and called as such:
* obj := Constructor();
* obj.Push(x);
* param_2 := obj.Pop();
* param_3 := obj.Top();
* param_4 := obj.Empty();
*/
```
javaScript:
使用数组（push, shift）模拟队列

23
problems/0232.用栈实现队列.md
View File

@ -205,33 +205,26 @@ class MyQueue:
def pop(self) -> int:
"""
1. 检查如果out里面元素，则直接pop
2. 如果out没有元素，就把in里面的元素（除了第一个）依次pop后装进out里面
3. 直接把in剩下的元素pop出来，就是queue头部的
Removes the element from in front of queue and returns that element.
"""
if self.empty:
if self.empty():
return None
if self.stack_out:
return self.stack_out.pop()
else:
for i in range(1, len(self.stack_in)):
for i in range(len(self.stack_in)):
self.stack_out.append(self.stack_in.pop())
return self.stack_in.pop()
return self.stack_out.pop()
def peek(self) -> int:
"""
1. 查out有没有元素，有就把最上面的返回
2. 如果out没有元素，就把in最下面的返回
Get the front element.
"""
if self.empty:
return None
if self.stack_out:
return self.stack_out[-1]
else:
return self.stack_in[0]
ans = self.pop()
self.stack_out.append(ans)
return ans
def empty(self) -> bool:

71
problems/0234.回文链表.md
View File

@ -144,6 +144,75 @@ public:
## Java
```java
// 方法一，使用数组
class Solution {
public boolean isPalindrome(ListNode head) {
int len = 0;
// 统计链表长度
ListNode cur = head;
while (cur != null) {
len++;
cur = cur.next;
}
cur = head;
int[] res = new int[len];
// 将元素加到数组之中
for (int i = 0; i < res.length; i++){
res[i] = cur.val;
cur = cur.next;
}
// 比较回文
for (int i = 0, j = len - 1; i < j; i++, j--){
if (res[i] != res[j]){
return false;
}
}
return true;
}
}
// 方法二，快慢指针
class Solution {
public boolean isPalindrome(ListNode head) {
// 如果为空或者仅有一个节点，返回true
if (head == null && head.next == null) return true;
ListNode slow = head;
ListNode fast = head;
ListNode pre = head;
while (fast != null && fast.next != null){
pre = slow; // 记录slow的前一个结点
slow = slow.next;
fast = fast.next.next;
}
pre.next = null; // 分割两个链表
// 前半部分
ListNode cur1 = head;
// 后半部分。这里使用了反转链表
ListNode cur2 = reverseList(slow);
while (cur1 != null){
if (cur1.val != cur2.val) return false;
// 注意要移动两个结点
cur1 = cur1.next;
cur2 = cur2.next;
}
return true;
}
ListNode reverseList(ListNode head){
// 反转链表
ListNode tmp = null;
ListNode pre = null;
while (head != null){
tmp = head.next;
head.next = pre;
pre = head;
head = tmp;
}
return pre;
}
}
```
## Python
@ -209,11 +278,13 @@ class Solution:
## Go
```go
```
## JavaScript
```js
```

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

73
problems/0347.前K个高频元素.md
View File

@ -189,6 +189,79 @@ class Solution:
Go：
```go
//方法一：小顶堆
func topKFrequent(nums []int, k int) []int {
map_num:=map[int]int{}
//记录每个元素出现的次数
for _,item:=range nums{
map_num[item]++
}
h:=&IHeap{}
heap.Init(h)
//所有元素入堆，堆的长度为k
for key,value:=range map_num{
heap.Push(h,[2]int{key,value})
if h.Len()>k{
heap.Pop(h)
}
}
res:=make([]int,k)
//按顺序返回堆中的元素
for i:=0;i<k;i++{
res[k-i-1]=heap.Pop(h).([2]int)[0]
}
return res
}
//构建小顶堆
type IHeap [][2]int
func (h IHeap) Len()int {
return len(h)
}
func (h IHeap) Less (i,j int) bool {
return h[i][1]<h[j][1]
}
func (h IHeap) Swap(i,j int) {
h[i],h[j]=h[j],h[i]
}
func (h *IHeap) Push(x interface{}){
*h=append(*h,x.([2]int))
}
func (h *IHeap) Pop() interface{}{
old:=*h
n:=len(old)
x:=old[n-1]
*h=old[0:n-1]
return x
}
//方法二:利用O(logn)排序
func topKFrequent(nums []int, k int) []int {
ans:=[]int{}
map_num:=map[int]int{}
for _,item:=range nums {
map_num[item]++
}
for key,_:=range map_num{
ans=append(ans,key)
}
//核心思想：排序
//可以不用包函数，自己实现快排
sort.Slice(ans,func (a,b int)bool{
return map_num[ans[a]]>map_num[ans[b]]
})
return ans[:k]
}
```
javaScript:
```js

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)

19
problems/0392.判断子序列.md
View File

@ -203,6 +203,25 @@ const isSubsequence = (s, t) => {
};
```
Go：
```go
func isSubsequence(s string, t string) bool {
dp := make([][]int,len(s)+1)
for i:=0;i<len(dp);i++{
dp[i] = make([]int,len(t)+1)
}
for i:=1;i<len(dp);i++{
for j:=1;j<len(dp[i]);j++{
if s[i-1] == t[j-1]{
dp[i][j] = dp[i-1][j-1] +1
}else{
dp[i][j] = dp[i][j-1]
}
}
}
return dp[len(s)][len(t)]==len(s)
}
```

26
problems/0404.左叶子之和.md
View File

@ -201,7 +201,31 @@ class Solution {
}
}
```
```java
// 层序遍历迭代法
class Solution {
public int sumOfLeftLeaves(TreeNode root) {
int sum = 0;
if (root == null) return 0;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
while (size -- > 0) {
TreeNode node = queue.poll();
if (node.left != null) { // 左节点不为空
queue.offer(node.left);
if (node.left.left == null && node.left.right == null){ // 左叶子节点
sum += node.left.val;
}
}
if (node.right != null) queue.offer(node.right);
}
}
return sum;
}
}
```
## Python

24
problems/0452.用最少数量的箭引爆气球.md
View File

@ -218,6 +218,30 @@ var findMinArrowShots = function(points) {
};
```
C:
```c
int cmp(const void *a,const void *b)
{
return ((*((int**)a))[0] > (*((int**)b))[0]);
}
int findMinArrowShots(int** points, int pointsSize, int* pointsColSize){
//将points数组作升序排序
qsort(points, pointsSize, sizeof(points[0]),cmp);
int arrowNum = 1;
int i = 1;
for(i = 1; i < pointsSize; i++) {
//若前一个气球与当前气球不重叠，证明需要增加箭的数量
if(points[i][0] > points[i-1][1])
arrowNum++;
else
//若前一个气球与当前气球重叠，判断并更新最小的x_end
points[i][1] = points[i][1] > points[i-1][1] ? points[i-1][1] : points[i][1];
}
return arrowNum;
}
```
-----------------------
* 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)

32
problems/0583.两个字符串的删除操作.md
View File

@ -147,8 +147,38 @@ class Solution:
```
Go：
```go
func minDistance(word1 string, word2 string) int {
dp := make([][]int, len(word1)+1)
for i := 0; i < len(dp); i++ {
dp[i] = make([]int, len(word2)+1)
}
//初始化
for i := 0; i < len(dp); i++ {
dp[i][0] = i
}
for j := 0; j < len(dp[0]); j++ {
dp[0][j] = j
}
for i := 1; i < len(dp); i++ {
for j := 1; j < len(dp[i]); j++ {
if word1[i-1] == word2[j-1] {
dp[i][j] = dp[i-1][j-1]
} else {
dp[i][j] = min(min(dp[i-1][j]+1, dp[i][j-1]+1), dp[i-1][j-1]+2)
}
}
}
return dp[len(dp)-1][len(dp[0])-1]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}
```
Javascript：
```javascript
const minDistance = (word1, word2) => {

32
problems/0704.二分查找.md
View File

@ -478,6 +478,38 @@ int search(int* nums, int numsSize, int target){
}
```
**PHP:**
```php
// 左闭右闭区间
class Solution {
/**
* @param Integer[] $nums
* @param Integer $target
* @return Integer
*/
function search($nums, $target) {
if (count($nums) == 0) {
return -1;
}
$left = 0;
$right = count($nums) - 1;
while ($left <= $right) {
$mid = floor(($left + $right) / 2);
if ($nums[$mid] == $target) {
return $mid;
}
if ($nums[$mid] > $target) {
$right = $mid - 1;
}
else {
$left = $mid + 1;
}
}
return -1;
}
}
```
-----------------------
* 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)

19
problems/0714.买卖股票的最佳时机含手续费（动态规划）.md
View File

@ -152,6 +152,25 @@ class Solution:
```
Go：
```Go
func maxProfit(prices []int, fee int) int {
n := len(prices)
dp := make([][2]int, n)
dp[0][0] = -prices[0]
for i := 1; i < n; i++ {
dp[i][1] = max(dp[i-1][1], dp[i-1][0]+prices[i]-fee)
dp[i][0] = max(dp[i-1][0], dp[i-1][1]-prices[i])
}
return dp[n-1][1]
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
```
Javascript：
```javascript

27
problems/0968.监控二叉树.md
View File

@ -368,7 +368,34 @@ class Solution:
return result
```
Go：
```go
const inf = math.MaxInt64 / 2
func minCameraCover(root *TreeNode) int {
var dfs func(*TreeNode) (a, b, c int)
dfs = func(node *TreeNode) (a, b, c int) {
if node == nil {
return inf, 0, 0
}
lefta, leftb, leftc := dfs(node.Left)
righta, rightb, rightc := dfs(node.Right)
a = leftc + rightc + 1
b = min(a, min(lefta+rightb, righta+leftb))
c = min(a, leftb+rightb)
return
}
_, ans, _ := dfs(root)
return ans
}
func min(a, b int) int {
if a <= b {
return a
}
return b
}
```
Javascript:
```Javascript
var minCameraCover = function(root) {

63
problems/0977.有序数组的平方.md
View File

@ -271,6 +271,69 @@ 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 {
/**
* @param Integer[] $nums
* @return Integer[]
*/
function sortedSquares($nums) {
// 双指针法
$res = [];
for ($i = 0; $i < count($nums); $i++) {
$res[$i] = 0;
}
$k = count($nums) - 1;
for ($i = 0, $j = count($nums) - 1; $i <= $j; ) {
if ($nums[$i] ** 2 < $nums[$j] ** 2) {
$res[$k--] = $nums[$j] ** 2;
$j--;
}
else {
$res[$k--] = $nums[$i] ** 2;
$i++;
}
}
return $res;
}
}
```
-----------------------
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

41
problems/1002.查找常用字符.md
View File

@ -268,6 +268,47 @@ func min(a,b int)int{
return a
}
```
Swift：
```swift
func commonChars(_ words: [String]) -> [String] {
var res = [String]()
if words.count < 1 {
return res
}
let aUnicodeScalarValue = "a".unicodeScalars.first!.value
let lettersMaxCount = 26
// 用于统计所有字符串每个字母出现的 最小 频率
var hash = Array(repeating: 0, count: lettersMaxCount)
// 统计第一个字符串每个字母出现的次数
for unicodeScalar in words.first!.unicodeScalars {
hash[Int(unicodeScalar.value - aUnicodeScalarValue)] += 1
}
// 统计除第一个字符串每个字母出现的次数
for idx in 1 ..< words.count {
var hashOtherStr = Array(repeating: 0, count: lettersMaxCount)
for unicodeScalar in words[idx].unicodeScalars {
hashOtherStr[Int(unicodeScalar.value - aUnicodeScalarValue)] += 1
}
// 更新hash,保证hash里统计的字母为出现的最小频率
for k in 0 ..< lettersMaxCount {
hash[k] = min(hash[k], hashOtherStr[k])
}
}
// 将hash统计的字符次数，转成输出形式
for i in 0 ..< lettersMaxCount {
while hash[i] != 0 { // 注意这里是while，多个重复的字符
let currentUnicodeScalarValue: UInt32 = UInt32(i) + aUnicodeScalarValue
let currentUnicodeScalar: UnicodeScalar = UnicodeScalar(currentUnicodeScalarValue)!
let outputStr = String(currentUnicodeScalar) // UnicodeScalar -> String
res.append(outputStr)
hash[i] -= 1
}
}
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)

9
problems/1005.K次取反后最大化的数组和.md
View File

@ -110,15 +110,16 @@ class Solution {
int len = nums.length;
for (int i = 0; i < len; i++) {
//从前向后遍历，遇到负数将其变为正数，同时K--
if (nums[i] < 0 && k > 0) {
if (nums[i] < 0 && K > 0) {
nums[i] = -nums[i];
k--;
K--;
}
}
// 如果K还大于0，那么反复转变数值最小的元素，将K用完
if (k % 2 == 1) nums[len - 1] = -nums[len - 1];
if (K % 2 == 1) nums[len - 1] = -nums[len - 1];
return Arrays.stream(nums).sum();
}
}
```

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！**

28
problems/剑指Offer58-II.左旋转字符串.md
View File

@ -214,6 +214,34 @@ var reverseLeftWords = function (s, n) {
};
```
Swift:
```swift
func reverseLeftWords(_ s: String, _ n: Int) -> String {
var ch = Array(s)
let len = ch.count
// 反转区间[0, n - 1]
reverseString(&ch, startIndex: 0, endIndex: n - 1)
// 反转区间[n, len - 1]
reverseString(&ch, startIndex: n, endIndex: len - 1)
// 反转区间[0, len - 1]，也就是整个字符串反转
reverseString(&ch, startIndex: 0, endIndex: len - 1)
return String(ch)
}
func reverseString(_ s: inout [Character], startIndex: Int, endIndex: Int) {
var start = startIndex
var end = endIndex
while start < end {
(s[start], s[end]) = (s[end], s[start])
start += 1
end -= 1
}
}
```