Merge branch 'master' of https://github.com/azl397985856/leetcode

Author: lucifer

problems/23.merge-k-sorted-lists.md

@@ -41,6 +41,10 @@ https://leetcode-cn.com/problems/merge-k-sorted-lists/description/
 
 ## 代码
 
+代码支持 JavaScript， Python3
+
+JavaScript Code：
+
 ```js
 /*
  * @lc app=leetcode id=23 lang=javascript
@@ -49,27 +53,6 @@ https://leetcode-cn.com/problems/merge-k-sorted-lists/description/
  *
  * https://leetcode.com/problems/merge-k-sorted-lists/description/
  *
- * algorithms
- * Hard (33.14%)
- * Total Accepted:    373.7K
- * Total Submissions: 1.1M
- * Testcase Example:  '[[1,4,5],[1,3,4],[2,6]]'
- *
- * Merge k sorted linked lists and return it as one sorted list. Analyze and
- * describe its complexity.
- *
- * Example:
- *
- *
- * Input:
- * [
- * 1->4->5,
- * 1->3->4,
- * 2->6
- * ]
- * Output: 1->1->2->3->4->4->5->6
- *
- *
  */
 function mergeTwoLists(l1, l2) {
   const dummyHead = {};
@@ -129,6 +112,51 @@ var mergeKLists = function(lists) {
 };
 ```
 
+Python3 Code：
+
+``` python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
+        n = len(lists)
+        
+        # basic cases
+        if lenth == 0: return None
+        if lenth == 1: return lists[0]
+        if lenth == 2: return self.mergeTwoLists(lists[0], lists[1])
+        
+        # divide and conqure if not basic cases
+        mid = n // 2
+        return self.mergeTwoLists(self.mergeKLists(lists[:mid]), self.mergeKLists(lists[mid:n]))
+    
+    
+    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
+        res = ListNode(0)
+        c1, c2, c3 = l1, l2, res
+        while c1 or c2:
+            if c1 and c2:
+                if c1.val < c2.val:
+                    c3.next = ListNode(c1.val)
+                    c1 = c1.next
+                else:
+                    c3.next = ListNode(c2.val)
+                    c2 = c2.next
+                c3 = c3.next
+            elif c1:
+                c3.next = c1
+                break
+            else:
+                c3.next = c2
+                break
+            
+        return res.next
+```
+
 ## 相关题目
 
 -[88.merge-sorted-array](./88.merge-sorted-array.md)

problems/42.trapping-rain-water.md

@@ -52,6 +52,10 @@ for(let i = 0; i < height.length; i++) {
 
 ## 代码
 
+代码支持 JavaScript，Python3:
+
+JavaScript Code:
+
 ```js
 
 /*
@@ -88,3 +92,30 @@ var trap = function(height) {
 };
 
 ```
+
+Python Code:
+
+```python
+
+class Solution:
+    def trap(self, height: List[int]) -> int:
+        maxLeft, maxRight, volum = 0, 0, 0
+        maxLeftStack, maxRightStack = [], []
+        for h in height:
+            if h > maxLeft:
+                maxLeftStack.append(h)
+                maxLeft = h
+            else:
+                maxLeftStack.append(maxLeft)
+        for h in height[::-1]:
+            if h > maxRight:
+                maxRightStack.append(h)
+                maxRight = h
+            else:
+                maxRightStack.append(maxRight)
+        maxRightStack = maxRightStack[::-1]
+        for i in range(1, len(height) - 1):
+            minSide = min(maxLeftStack[i], maxRightStack[i])
+            volum += minSide - height[i]
+        return volum      
+```

problems/445.add-two-numbers-ii.md

@@ -37,7 +37,7 @@ Output: 7 -> 8 -> 0 -> 7
 
 ## 代码
 
-- 语言支持：JS，C++
+- 语言支持：JS，C++, Python3
 
 JavaScript Code:
 
@@ -210,3 +210,50 @@ private:
     }
 };
 ```
+
+Python Code:
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
+        def listToStack(l: ListNode) -> list:
+            stack, c = [], l
+            while c:
+                stack.append(c.val)
+                c = c.next
+            return stack
+            
+        # transfer l1 and l2 into stacks
+        stack1, stack2 = listToStack(l1), listToStack(l2)
+            
+        # add stack1 and stack2
+        diff = abs(len(stack1) - len(stack2))
+        stack1 = ([0]*diff + stack1 if len(stack1) < len(stack2) else stack1)
+        stack2 = ([0]*diff + stack2 if len(stack2) < len(stack1) else stack2)
+        stack3 = [x + y for x, y in zip(stack1, stack2)]
+        
+        # calculate carry for each item in stack3 and add one to the item before it
+        carry = 0
+        for i, val in enumerate(stack3[::-1]):
+            index = len(stack3) - i - 1
+            carry, stack3[index] = divmod(val + carry, 10)
+            if carry and index == 0: 
+                stack3 = [1] + stack3
+            elif carry:
+                stack3[index - 1] += 1
+        
+        # transfer stack3 to a linkedList
+        result = ListNode(0)
+        c = result
+        for item in stack3:
+            c.next = ListNode(item)
+            c = c.next
+        
+        return result.next
+```

problems/73.set-matrix-zeroes.md

@@ -179,6 +179,48 @@ var setZeroes = function(matrix) {
 };
 ```
 Python3 Code:
+
+直接修改第一行和第一列为0的解法：
+```python
+class Solution:
+    def setZeroes(self, matrix: List[List[int]]) -> None:
+        """
+        Do not return anything, modify matrix in-place instead.
+        """
+        def setRowZeros(matrix: List[List[int]], i:int) -> None:
+            C = len(matrix[0])
+            matrix[i] = [0] * C
+        
+        def setColZeros(matrix: List[List[int]], j:int) -> None:
+            R = len(matrix)
+            for i in range(R):
+                matrix[i][j] = 0
+            
+        isCol = False
+        R = len(matrix)
+        C = len(matrix[0])
+        
+        for i in range(R):
+            if matrix[i][0] == 0:
+                isCol = True
+            for j in range(1, C):
+                if matrix[i][j] == 0:
+                    matrix[i][0] = 0
+                    matrix[0][j] = 0
+        for j in range(1, C):
+            if matrix[0][j] == 0:
+                setColZeros(matrix, j)
+                
+        for i in range(R):
+            if matrix[i][0] == 0:
+                setRowZeros(matrix, i)
+        
+        if isCol:
+            setColZeros(matrix, 0)
+
+```
+
+另一种方法是用一个特殊符合标记需要改变的结果，只要这个特殊标记不在我们的题目数据范围（0和1）即可，这里用None。
 ```python
 class Solution:
     def setZeroes(self, matrix: List[List[int]]) -> None: