diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index f8c9da5f..fd17af62 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -9,7 +9,7 @@
 
 ## 1. 两数之和
 
-https://leetcode-cn.com/problems/two-sum/
+[力扣题目链接](https://leetcode-cn.com/problems/two-sum/)
 
 给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
 
@@ -29,10 +29,10 @@ https://leetcode-cn.com/problems/two-sum/
 很明显暴力的解法是两层for循环查找，时间复杂度是O(n^2)。
 
 建议大家做这道题目之前，先做一下这两道
-* [242. 有效的字母异位词](https://mp.weixin.qq.com/s/ffS8jaVFNUWyfn_8T31IdA)
-* [349. 两个数组的交集](https://mp.weixin.qq.com/s/aMSA5zrp3jJcLjuSB0Es2Q)
+* [242. 有效的字母异位词](https://www.programmercarl.com/0242.有效的字母异位词.html)
+* [349. 两个数组的交集](https://www.programmercarl.com/0349.两个数组的交集.html)
 
-[242. 有效的字母异位词](https://mp.weixin.qq.com/s/ffS8jaVFNUWyfn_8T31IdA) 这道题目是用数组作为哈希表来解决哈希问题，[349. 两个数组的交集](https://mp.weixin.qq.com/s/aMSA5zrp3jJcLjuSB0Es2Q)这道题目是通过set作为哈希表来解决哈希问题。
+[242. 有效的字母异位词](https://www.programmercarl.com/0242.有效的字母异位词.html) 这道题目是用数组作为哈希表来解决哈希问题，[349. 两个数组的交集](https://www.programmercarl.com/0349.两个数组的交集.html)这道题目是通过set作为哈希表来解决哈希问题。
 
 本题呢，则要使用map，那么来看一下使用数组和set来做哈希法的局限。
 
@@ -51,7 +51,7 @@ C++中map，有三种类型：
 
 std::unordered_map 底层实现为哈希表，std::map 和std::multimap 的底层实现是红黑树。
 
-同理，std::map 和std::multimap 的key也是有序的（这个问题也经常作为面试题，考察对语言容器底层的理解）。 更多哈希表的理论知识请看[关于哈希表，你该了解这些！](https://mp.weixin.qq.com/s/RSUANESA_tkhKhYe3ZR8Jg)。
+同理，std::map 和std::multimap 的key也是有序的（这个问题也经常作为面试题，考察对语言容器底层的理解）。 更多哈希表的理论知识请看[关于哈希表，你该了解这些！](https://www.programmercarl.com/哈希表理论基础.html)。
 
 **这道题目中并不需要key有序，选择std::unordered_map 效率更高！**
 
@@ -110,13 +110,14 @@ Python：
 ```python
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
-        hashmap={}
-        for ind,num in enumerate(nums):
-            hashmap[num] = ind
-        for i,num in enumerate(nums):
-            j = hashmap.get(target - num)
-            if j is not None and i!=j:
-                return [i,j]
+        records = dict()
+
+        # 用枚举更方便，就不需要通过索引再去取当前位置的值
+        for idx, val in enumerate(nums):
+            if target - val not in records:
+                records[val] = idx
+            else:
+                return [records[target - val], idx] # 如果存在就返回字典记录索引和当前索引
 ```
 
 
diff --git a/problems/0059.螺旋矩阵II.md b/problems/0059.螺旋矩阵II.md
index e46dae6d..6df8c83d 100644
--- a/problems/0059.螺旋矩阵II.md
+++ b/problems/0059.螺旋矩阵II.md
@@ -302,6 +302,61 @@ func generateMatrix(n int) [][]int {
 }
 ```
 
+Swift:
+
+```swift
+func generateMatrix(_ n: Int) -> [[Int]] {
+    var result = [[Int]](repeating: [Int](repeating: 0, count: n), count: n)
+
+    var startRow = 0
+    var startColumn = 0
+    var loopCount = n / 2
+    let mid = n / 2
+    var count = 1
+    var offset = 1
+    var row: Int
+    var column: Int
+
+    while loopCount > 0 {
+        row = startRow
+        column = startColumn
+
+        for c in column ..< startColumn + n - offset {
+            result[startRow][c] = count
+            count += 1
+            column += 1
+        }
+
+        for r in row ..< startRow + n - offset {
+            result[r][column] = count
+            count += 1
+            row += 1
+        }
+
+        for _ in startColumn ..< column {
+            result[row][column] = count
+            count += 1
+            column -= 1
+        }
+
+        for _ in startRow ..< row {
+            result[row][column] = count
+            count += 1
+            row -= 1
+        }
+
+        startRow += 1
+        startColumn += 1
+        offset += 2
+        loopCount -= 1
+    }
+
+    if (n % 2) != 0 {
+        result[mid][mid] = count
+    }
+    return result
+}
+```
 
 
 
diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md
index b37270e2..17642793 100644
--- a/problems/0129.求根到叶子节点数字之和.md
+++ b/problems/0129.求根到叶子节点数字之和.md
@@ -165,7 +165,32 @@ public:
 Java： 
 
 Python：
+```python3
+class Solution:
+    def sumNumbers(self, root: TreeNode) -> int:
+        res = 0
+        path = []
+        def backtrace(root):
+            nonlocal res
+            if not root: return # 节点空则返回
+            path.append(root.val)
+            if not root.left and not root.right: # 遇到了叶子节点
+                res += get_sum(path)
+            if root.left: # 左子树不空
+                backtrace(root.left)
+            if root.right: # 右子树不空
+                backtrace(root.right)
+            path.pop()
 
+        def get_sum(arr):
+            s = 0
+            for i in range(len(arr)):
+                s = s * 10 + arr[i]
+            return s
+
+        backtrace(root)
+        return res
+```
 Go：
 
 JavaScript：
diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md
index 62232051..76df63b7 100644
--- a/problems/0143.重排链表.md
+++ b/problems/0143.重排链表.md
@@ -222,7 +222,61 @@ public class ReorderList {
 ```
 
 Python：
+```python3
+# 方法二 双向队列
+class Solution:
+    def reorderList(self, head: ListNode) -> None:
+        """
+        Do not return anything, modify head in-place instead.
+        """
+        d = collections.deque()
+        tmp = head
+        while tmp.next: # 链表除了首元素全部加入双向队列
+            d.append(tmp.next)
+            tmp = tmp.next
+        tmp = head
+        while len(d): # 一后一前加入链表
+            tmp.next = d.pop()
+            tmp = tmp.next
+            if len(d):
+                tmp.next = d.popleft()
+                tmp = tmp.next
+        tmp.next = None # 尾部置空
+ 
+# 方法三 反转链表
+class Solution:
+    def reorderList(self, head: ListNode) -> None:
+        if head == None or head.next == None:
+            return True
+        slow, fast = head, head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+        right = slow.next # 分割右半边
+        slow.next = None # 切断
+        right = self.reverseList(right) #反转右半边
+        left = head
+        # 左半边一定比右半边长, 因此判断右半边即可
+        while right:
+            curLeft = left.next
+            left.next = right
+            left = curLeft
 
+            curRight = right.next
+            right.next = left
+            right = curRight
+
+
+    def reverseList(self, head: ListNode) -> ListNode:
+        cur = head   
+        pre = None
+        while(cur!=None):
+            temp = cur.next # 保存一下cur的下一个节点
+            cur.next = pre # 反转
+            pre = cur
+            cur = temp
+        return pre
+```
 Go：
 
 JavaScript：
diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md
index 1c630b6a..2e784e6a 100644
--- a/problems/0202.快乐数.md
+++ b/problems/0202.快乐数.md
@@ -111,25 +111,29 @@ Python：
 ```python
 class Solution:
     def isHappy(self, n: int) -> bool:
-        set_ = set()
-        while 1:
-            sum_ = self.getSum(n)
-            if sum_ == 1:
+        def calculate_happy(num):
+            sum_ = 0
+            
+            # 从个位开始依次取，平方求和
+            while num:
+                sum_ += (num % 10) ** 2
+                num = num // 10
+            return sum_
+
+        # 记录中间结果
+        record = set()
+
+        while True:
+            n = calculate_happy(n)
+            if n == 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，立刻return false
-            if sum_ in set_:
+            
+            # 如果中间结果重复出现，说明陷入死循环了，该数不是快乐数
+            if n in record:
                 return False
             else:
-                set_.add(sum_)
-            n = sum_
-            
-    #取数值各个位上的单数之和
-    def getSum(self, n):
-        sum_ = 0
-        while n > 0:
-            sum_ += (n%10) * (n%10)
-            n //= 10
-        return sum_
+                record.add(n)
+
 ```
 
 Go：
diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md
index a2c6e90d..9235d47e 100644
--- a/problems/0203.移除链表元素.md
+++ b/problems/0203.移除链表元素.md
@@ -304,6 +304,34 @@ var removeElements = function(head, val) {
 };
 ```
 
+Swift:
+
+```swift
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     public var val: Int
+ *     public var next: ListNode?
+ *     public init() { self.val = 0; self.next = nil; }
+ *     public init(_ val: Int) { self.val = val; self.next = nil; }
+ *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
+ * }
+ */
+func removeElements(_ head: ListNode?, _ val: Int) -> ListNode? {
+    let dummyNode = ListNode()
+    dummyNode.next = head
+    var currentNode = dummyNode
+    while let curNext = currentNode.next {
+        if curNext.val == val {
+            currentNode.next = curNext.next
+        } else {
+            currentNode = curNext
+        }
+    }
+    return dummyNode.next
+}
+```
+
 
 
 
diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md
index 6a24b1d0..b3ad899c 100644
--- a/problems/0234.回文链表.md
+++ b/problems/0234.回文链表.md
@@ -148,7 +148,62 @@ public:
 
 ## Python
 
-```python
+```python3
+#数组模拟
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        length = 0
+        tmp = head
+        while tmp: #求链表长度
+            length += 1
+            tmp = tmp.next
+        
+        result = [0] * length
+        tmp = head
+        index = 0
+        while tmp: #链表元素加入数组
+            result[index] = tmp.val
+            index += 1
+            tmp = tmp.next
+        
+        i, j = 0, length - 1
+        while i < j: # 判断回文
+            if result[i] != result[j]:
+                return False
+            i += 1
+            j -= 1
+        return True
+        
+#反转后半部分链表
+class Solution:
+    def isPalindrome(self, head: ListNode) -> bool:
+        if head == None or head.next == None:
+            return True
+        slow, fast = head, head
+        while fast and fast.next:
+            pre = slow
+            slow = slow.next
+            fast = fast.next.next
+        
+        pre.next = None # 分割链表
+        cur1 = head # 前半部分
+        cur2 = self.reverseList(slow) # 反转后半部分，总链表长度如果是奇数，cur2比cur1多一个节点
+        while cur1:
+            if cur1.val != cur2.val:
+                return False
+            cur1 = cur1.next
+            cur2 = cur2.next
+        return True
+
+    def reverseList(self, head: ListNode) -> ListNode:
+        cur = head   
+        pre = None
+        while(cur!=None):
+            temp = cur.next # 保存一下cur的下一个节点
+            cur.next = pre # 反转
+            pre = cur
+            cur = temp
+        return pre
 ```
 
 ## Go
diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md
index 29c1c144..7489352d 100644
--- a/problems/0349.两个数组的交集.md
+++ b/problems/0349.两个数组的交集.md
@@ -121,13 +121,7 @@ Python：
 ```python
 class Solution:
     def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
-        result_set = set()
-        
-        set1 = set(nums1)
-        for num in nums2:
-            if num in set1:
-                result_set.add(num) # set1里出现的nums2元素 存放到结果
-        return list(result_set)
+        return list(set(nums1) & set(nums2))    # 两个数组先变成集合，求交集后还原为数组
 ```
 
 
diff --git a/problems/0724.寻找数组的中心索引.md b/problems/0724.寻找数组的中心索引.md
index 3ed68d47..b4115893 100644
--- a/problems/0724.寻找数组的中心索引.md
+++ b/problems/0724.寻找数组的中心索引.md
@@ -89,7 +89,16 @@ class Solution {
 
 ## Python
 
-```python
+```python3
+class Solution:
+    def pivotIndex(self, nums: List[int]) -> int:
+        numSum = sum(nums) #数组总和
+        leftSum = 0
+        for i in range(len(nums)):
+            if numSum - leftSum -nums[i] == leftSum: #左右和相等 
+                return i
+            leftSum += nums[i]
+        return -1
 ```
 
 ## Go
diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md
index 92db204d..97d7091e 100644
--- a/problems/0922.按奇偶排序数组II.md
+++ b/problems/0922.按奇偶排序数组II.md
@@ -149,7 +149,32 @@ class Solution {
 
 ## Python
 
-```python
+```python3
+#方法2
+class Solution:
+    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
+        result = [0]*len(nums)
+        evenIndex = 0
+        oddIndex = 1
+        for i in range(len(nums)):
+            if nums[i] % 2: #奇数
+                result[oddIndex] = nums[i]
+                oddIndex += 2
+            else: #偶数
+                result[evenIndex] = nums[i]
+                evenIndex += 2
+        return result
+
+#方法3
+class Solution:
+    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
+        oddIndex = 1
+        for i in range(0,len(nums),2): #步长为2
+            if nums[i] % 2: #偶数位遇到奇数
+                while  nums[oddIndex] % 2: #奇数位找偶数
+                    oddIndex += 2
+                nums[i], nums[oddIndex] = nums[oddIndex], nums[i]
+        return nums
 ```
 
 ## Go
diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md
index 9e71ec0d..71c46401 100644
--- a/problems/0977.有序数组的平方.md
+++ b/problems/0977.有序数组的平方.md
@@ -252,6 +252,24 @@ func sortedSquares(_ nums: [Int]) -> [Int] {
 }
 ```
 
+Ruby:
+
+```ruby
+def sorted_squares(nums)
+  left, right, result = 0, nums.size - 1, []
+  while left <= right
+    if nums[left]**2 > nums[right]**2
+      result << nums[left]**2
+      left += 1
+    else
+      result << nums[right]**2
+      right -= 1
+    end
+  end
+  result.reverse
+end
+```
+
 
 
 -----------------------
diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 9acda71c..8c3a5831 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -160,34 +160,21 @@ Python：
 
 class Solution:
     def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
-        lengthA,lengthB = 0,0
-        curA,curB = headA,headB
-        while(curA!=None): #求链表A的长度
-            curA = curA.next
-            lengthA +=1
-        
-        while(curB!=None): #求链表B的长度
-            curB = curB.next
-            lengthB +=1
-        
-        curA, curB = headA, headB
+        """
+        根据快慢法则，走的快的一定会追上走得慢的。
+        在这道题里，有的链表短，他走完了就去走另一条链表，我们可以理解为走的快的指针。
 
-        if lengthB>lengthA: #让curA为最长链表的头，lenA为其长度
-            lengthA, lengthB = lengthB, lengthA
-            curA, curB = curB, curA
+        那么，只要其中一个链表走完了，就去走另一条链表的路。如果有交点，他们最终一定会在同一个
+        位置相遇
+        """
+        cur_a, cur_b = headA, headB     # 用两个指针代替a和b
 
-        gap = lengthA - lengthB #求长度差
-        while(gap!=0): 
-            curA = curA.next #让curA和curB在同一起点上
-            gap -= 1
         
-        while(curA!=None):
-            if curA == curB:
-                return curA
-            else:
-                curA = curA.next
-                curB = curB.next
-        return None
+        while cur_a != cur_b:
+            cur_a = cur_a.next if cur_a else headB      # 如果a走完了，那么就切换到b走
+            cur_b = cur_b.next if cur_b else headA      # 同理，b走完了就切换到a
+        
+        return cur_a
 ```
 
 Go：