Merge pull request #1099 from Verolilo/master

更新0070.爬楼梯.md python版本

problems/0070.爬楼梯.md

@@ -256,16 +256,28 @@ public int climbStairs(int n) {
 ### Python
 
 ```python
+# 空间复杂度为O(n)版本
 class Solution:
     def climbStairs(self, n: int) -> int:
-        # dp[i]表示爬到第i级楼梯的种数， (1, 2) (2, 1)是两种不同的类型
+        # dp[i] 为第 i 阶楼梯有多少种方法爬到楼顶
-        dp = [0] * (n + 1)
+        dp=[0]*(n+1)
-        dp[0] = 1
+        dp[0]=1
-        for i in range(n+1):
+        dp[1]=1
-            for j in range(1, 3):
+        for i in range(2,n+1):
-                if i>=j:
+            dp[i]=dp[i-1]+dp[i-2]
-                    dp[i] += dp[i-j]
+        return dp[n]
-        return dp[-1]
+        
+# 空间复杂度为O(1)版本
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        dp=[0]*(n+1)
+        dp[0]=1
+        dp[1]=1
+        for i in range(2,n+1):
+            tmp=dp[0]+dp[1]
+            dp[0]=dp[1]
+            dp[1]=tmp
+        return dp[1]
 ```
 
 ### Go 

problems/0213.打家劫舍II.md

@@ -123,22 +123,24 @@ Python：
 ```Python
 class Solution:
     def rob(self, nums: List[int]) -> int:
-      if (n := len(nums)) == 0:
+        #在198入门级的打家劫舍问题上分两种情况考虑
-        return 0
+        #一是不偷第一间房，二是不偷最后一间房
-      if n == 1:
+        if len(nums)==1:#题目中提示nums.length>=1,所以不需要考虑len(nums)==0的情况
-        return nums[0]
+            return nums[0]
-      result1 = self.robRange(nums, 0, n - 2)
+        val1=self.roblist(nums[1:])#不偷第一间房
-      result2 = self.robRange(nums, 1, n - 1)
+        val2=self.roblist(nums[:-1])#不偷最后一间房
-      return max(result1 , result2)
+        return max(val1,val2)
 
-    def robRange(self, nums: List[int], start: int, end: int) -> int:
+    def robRange(self,nums):
-      if end == start: return nums[start]
+        l=len(nums)
-      dp = [0] * len(nums)
+        dp=[0]*l
-      dp[start] = nums[start]
+        dp[0]=nums[0]
-      dp[start + 1] = max(nums[start], nums[start + 1])
+        for i in range(1,l):
-      for i in range(start + 2, end + 1):
+            if i==1:
-        dp[i] = max(dp[i -2] + nums[i], dp[i - 1])
+                dp[i]=max(dp[i-1],nums[i])
-      return dp[end]
+            else:
+                dp[i]=max(dp[i-1],dp[i-2]+nums[i])
+        return dp[-1]
 ```
 
 javascipt: