Change link of 1290 and 1342. Fix code style

Author: qiyuangong

README.md

@@ -222,9 +222,9 @@ I'm currently working on [Analytics-Zoo](https://github.com/intel-analytics/anal
 | 1089 | [Duplicate Zeros](https://leetcode.com/problems/duplicate-zeros/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1089_Duplicate_Zeros.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1089_Duplicate_Zeros.java) | 2 Pass, store last position and final move steps, O(n) and O(1) |
 | 1108 | [Defanging an IP Address](https://leetcode.com/problems/defanging-an-ip-address/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1108_Defanging_an_IP_Address.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1108_Defanging_an_IP_Address.java) | String manipulate (split, replace and join), O(n) and O(n) |
 | 1260 | [Shift 2D Grid](https://leetcode.com/problems/shift-2d-grid/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1260_Shift_2D_Grid.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1260_Shift_2D_Grid.java) | Final position of each element can be computed according to k, m and n, e.g., k == mn, then don't move, O(mn) and O(mn) |
-| 1290 | [Convert Binary Number in a Linked List to Integer](https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/) | [Python](https://github.com/srivastavaayu/leetcode-1/blob/master/python/1290_Convert_Binary_Number_in_a_Linked_List_to_Integer.py) | Take 2 to the power digit position from right (starting from 0) and multiply it with the digit |
+| 1290 | [Convert Binary Number in a Linked List to Integer](https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1290_Convert_Binary_Number_in_a_Linked_List_to_Integer.py) | Take 2 to the power digit position from right (starting from 0) and multiply it with the digit |
 | 1337 | [The K Weakest Rows in a Matrix](https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1337_The_K_Weakest_Rows_in_a_Matrix.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1337_The_K_Weakest_Rows_in_a_Matrix.java) | Check by row, from left to right, until encount first zero, O(mn) and O(1) |
-| 1342 | [Number of Steps to Reduce a Number to Zero](https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/) | [Python](https://github.com/srivastavaayu/leetcode-1/blob/master/python/1342_Number_of_Steps_to_Reduce_a_Number_to_Zero.py) | If number is divisible by 2, divide the number by 2, else subtract 1 from the number, and output the number of steps |
+| 1342 | [Number of Steps to Reduce a Number to Zero](https://leetcode.com/problems/number-of-steps-to-reduce-a-number-to-zero/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1342_Number_of_Steps_to_Reduce_a_Number_to_Zero.py) | If number is divisible by 2, divide the number by 2, else subtract 1 from the number, and output the number of steps, O(logn) and O(1) |
 | 1365 | [How Many Numbers Are Smaller Than the Current Number](https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1365_How_Many_Numbers_Are_Smaller_Than_the_Current_Number.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1365_How_Many_Numbers_Are_Smaller_Than_the_Current_Number.java) | 1. Sort and get position in sorted nums, O(nlogn) and O(n)<br>2. Fill count into 0-100, O(n) and O(1) |
 | 1480 | [Running Sum of 1d Array](https://leetcode.com/problems/running-sum-of-1d-array/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/1480_Running_Sum_of_1d_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/1480_Running_Sum_of_1d_Array.java) | 1. Go through the array, O(n) and O(1)<br>2. Accumulate API |
 

python/1290_Convert_Binary_Number_in_a_Linked_List_to_Integer.py

@@ -33,15 +33,16 @@
 
 class Solution:
     def getDecimalValue(self, head: ListNode) -> int:
-        binary_numbers_list=[]
+        binary_numbers_list = []
         binary_numbers_list.append(head.val)
-        while(head.next!=None):
-            head=head.next
+        while(head.next is not None):
+            head = head.next
             binary_numbers_list.append(head.val)
-        answer=0
-        power=0
-        for digit in range(len(binary_numbers_list)-1,-1,-1):
-            if(binary_numbers_list[digit]>0):
-                answer+=((2**power)*binary_numbers_list[digit])
-            power+=1
+        answer = 0
+        power = 0
+        # from len(binary_numbers_list) - 1 -> 0
+        for digit in range(len(binary_numbers_list) - 1, -1, -1):
+            if(binary_numbers_list[digit] > 0):
+                answer += ((2 ** power) * binary_numbers_list[digit])
+            power += 1
         return answer

python/1342_Number_of_Steps_to_Reduce_a_Number_to_Zero.py

@@ -32,13 +32,27 @@
 
 class Solution:
     def numberOfSteps (self, num: int) -> int:
-        steps=0
-        while(num>0):
-            if(num%2==0):
-                num=num/2
-                steps+=1
+        steps = 0
+        while(num > 0):
+            if(num % 2 == 0):
+                num = num / 2
+                steps + =1
             else:
-                num=num-1
-                steps+=1
+                num = num - 1
+                steps += 1
         return steps
-        
+
+    # def numberOfSteps (self, num: int) -> int:
+    #     ans = 0
+    #     # check the number of 1
+    #     while num:
+    #         ans += (num & 1) + 1
+    #         num >>= 1
+    #     return ans - 1
+
+    # def numberOfSteps (self, num: int) -> int:
+    #     ones, zeros = self.bitCount(num)
+    #     if ones == 0:
+    #         return 0
+    #     else:
+    #         return 2 * ones + zeros - 1