diff --git a/Dynamic Program/Kilien/121.best-time-to-buy-and-sell-stock.py b/Dynamic Program/Kilien/121.best-time-to-buy-and-sell-stock.py
new file mode 100644
index 0000000000000000000000000000000000000000..90ddc22b9c3aa9c4480f66f50d789d27c33cb658
--- /dev/null
+++ b/Dynamic Program/Kilien/121.best-time-to-buy-and-sell-stock.py	
@@ -0,0 +1,37 @@
+# @Author：Kilien
+# @lc app=leetcode id=121 lang=python3
+# [121] Best Time to Buy and Sell Stock
+# 思路：DP[i][j]，两个维度，i = 天数，j = 是否持有股票;当j=0，只能买股票，j=1，可以卖股票
+# maxProfit[i, 0] = max(maxProfit[i-1][0], maxProfit[i-1][1] + prices[i]) 
+# maxProfit[i, 1] = max(maxProfit[i-1][1], maxProfit[i-1][0] - prices[i])
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        if not prices: return 0
+
+        res = 0
+        #三种状态：
+        #profit[0][0] 未持有股票
+        #profit[0][1] 买入股票
+        #profit[0][2] 卖出股票
+        profit = [[0 for i in range(3)] for i in range(len(prices))]
+        profit[0][0], profit[0][1], profit[0][2] = 0, -prices[0], 0
+
+        for i in range(1, len(prices)):
+            profit[i][0] = profit[i-1][0]
+            profit[i][1] = max(profit[i-1][1], profit[i-1][0] - prices[i])
+            profit[i][2] = profit[i-1][1] + prices[i]
+            res = max(res, profit[i][0], profit[i][1], profit[i][2])
+        return res
+'''
+# 思路：由于只能买卖一次股票，可以遍历一次股票，将最近的股票交易利润记录下来，找出最大利润
+# maxCur = current maximum value
+# maxSoFar = maximum value found so far
+
+    def maxProfit(self, prices: List[int]) -> int:
+        maxCur, maxSoFar = 0, 0
+        for i in range(1, len(prices)):
+            maxCur += prices[i] - prices[i-1]
+            maxCur = max(0, maxCur)
+            maxSoFar = max(maxCur, maxSoFar)
+        return maxSoFar
+'''
diff --git a/Dynamic Program/Kilien/122.best-time-to-buy-and-sell-stock-ii.py b/Dynamic Program/Kilien/122.best-time-to-buy-and-sell-stock-ii.py
new file mode 100644
index 0000000000000000000000000000000000000000..0f3d1ea3aa71d3ad9d077e6accd7e1d9e0098ecc
--- /dev/null
+++ b/Dynamic Program/Kilien/122.best-time-to-buy-and-sell-stock-ii.py	
@@ -0,0 +1,25 @@
+# @Author:Kilien
+# @lc app=leetcode id=122 lang=python3
+#
+# [122] Best Time to Buy and Sell Stock II
+#
+# https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
+# 思路：两个状态：是否持有股票；
+# 状态转换：
+# hold -> do nothing -> hold
+# hold -> sell -> not hold
+# not hold -> do nothing -> not hold
+# not hold -> buy -> hold
+class Solution:
+    def maxProfit(self, prices: List[int]) -> int:
+        hold = float('-inf')#负无穷大
+        notHold = 0
+        for i in prices:
+            hold = max(hold, notHold - i)
+            notHold = max(notHold, hold + i)
+        return max(hold, notHold)
+'''
+# 思路：明天的股价高于今天，即今天买入，明天卖出；反之，不交易
+    def maxProfit(self, prices: List[int]) -> int:
+        return sum(max(prices[i + 1] - prices[i], 0) for i in range(len(prices) - 1))
+'''
diff --git a/Dynamic Program/Kilien/322.coin-change.py b/Dynamic Program/Kilien/322.coin-change.py
new file mode 100644
index 0000000000000000000000000000000000000000..ed58add0d83f4e4d251e7d1e3e3422601e8c311d
--- /dev/null
+++ b/Dynamic Program/Kilien/322.coin-change.py	
@@ -0,0 +1,36 @@
+# @Author：Kilien
+# @lc app=leetcode id=322 lang=python3
+#
+# [322] Coin Change
+#
+# https://leetcode.com/problems/coin-change/description/
+
+'''
+ dp[i]是组成目标金额所需的最少的硬币数量，coin指代不同面值的硬币；
+ dp[amount] * (amount + 1) 初始化数组；dp[0]作为哨兵节点，初始状态
+ 两层循环，先遍历金额，后遍历硬币面值；将最小组合动态置入dp状态数组
+ dp方程：dp[i] = min(dp[i - coin]) + 1
+'''
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        dp = [amount] * (amount + 1)
+        dp[0] = 0
+        for i in range(1, amount + 1):
+            for c in coins:
+                if i >= c:
+                    dp[i] = min(dp[i], dp[i - c] + 1)
+        
+        if dp[amount] == amount + 1:
+            return -1
+        return dp[amount]
+
+'''’
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        if amount == 0:return 0
+        Max = float('inf')
+        dp = [0] + [Max] * amount
+
+        for i in range(1, amount + 1):
+            dp[i] = min([dp[i - c] if i - c >= 0 else Max for c in coins]) + 1
+        return [dp[amount], -1][dp[amount] == Max]
+'''
diff --git a/Recursion/Kilien/46.permutations.py b/Recursion/Kilien/46.permutations.py
new file mode 100644
index 0000000000000000000000000000000000000000..26f7e4a4651391dbd5bd821fe8647a635bec7137
--- /dev/null
+++ b/Recursion/Kilien/46.permutations.py
@@ -0,0 +1,40 @@
+# @Author：Kilien
+# @lc app=leetcode id=46 lang=python3
+#
+# [46] Permutations
+#
+class Solution:
+'''
+  构造递归函数:
+  终止条件：当 start == end 则将该组合放入list
+  遍历 nums 数组，顺序置换两位数字位置； 
+  start + 1，递归
+  还原置换的数字位置
+'''
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        def backtrack(start, end):
+            if start == end:
+                ans.append(nums[:])
+            for i in range(start, end):
+                nums[start], nums[i] = nums[i], nums[start]
+                backtrack(start+1, end)
+                nums[start], nums[i] = nums[i], nums[start]
+                
+        ans = []
+        backtrack(0, len(nums))
+        return ans
+
+'''
+# DFS
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        res = []
+        self.dfs(nums, [], res)
+        return res 
+
+    def dfs(self, nums, path, res):
+        if not nums:
+            res.append(path)
+            return # backtracking
+        for i in range(len(nums)):
+            self.dfs(nums[:i]+nums[i+1:], path+[nums[i]], res)
+'''
diff --git a/Recursion/Kilien/69.sqrt-x.py b/Recursion/Kilien/69.sqrt-x.py
new file mode 100644
index 0000000000000000000000000000000000000000..a033d6c11191f180085c084966134b5785ec8798
--- /dev/null
+++ b/Recursion/Kilien/69.sqrt-x.py
@@ -0,0 +1,30 @@
+# Author：Kilien
+# @lc app=leetcode id=69 lang=python3
+#
+# [69] Sqrt(x)
+#
+class Solution:
+# 牛顿整除法
+    def mySqrt(self, x: int) -> int:
+        r = x + 1
+        while r*r > x:
+            r = int(r - (r*r -x)/(2*r))
+        return r
+        
+'''
+# 强制转换。。
+    def mySqrt(self, x: int) -> int:
+        return int(x**0.5)
+
+# 思路：二分查找
+    def mySqrt(self, x: int) -> int:
+        l, r = 0, x
+        while l <= r:
+            mid = l + (r-l)//2
+            if mid * mid <= x < (mid+1)*(mid+1):
+                return mid
+            elif x < mid * mid:
+                r = mid
+            else:
+                l = mid + 1   
+'''