Merge pull request #402 from borninfreedom/master

update 四数相加II，添加了一个更加简洁的python代码

problems/0001.两数之和.md

@@ -107,7 +107,7 @@ public int[] twoSum(int[] nums, int target) {
 
 Python：
 
-```python3
+```python
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
         hashmap={}

problems/0349.两个数组的交集.md

@@ -118,7 +118,7 @@ class Solution {
 ```
 
 Python：
-```python3
+```python
 class Solution:
     def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
         result_set = set()

problems/0383.赎金信.md

@@ -135,8 +135,52 @@ class Solution {
 
 ```
 
-Python：
-```py
+Python写法一（使用数组作为哈希表）：
+
+```python
+class Solution:
+    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
+
+        arr = [0] * 26
+
+        for x in magazine:
+            arr[ord(x) - ord('a')] += 1
+
+        for x in ransomNote:
+            if arr[ord(x) - ord('a')] == 0:
+                return False
+            else:
+                arr[ord(x) - ord('a')] -= 1
+        
+        return True
+```
+
+Python写法二（使用defaultdict）：
+
+```python
+class Solution:
+    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
+
+        from collections import defaultdict
+
+        hashmap = defaultdict(int)
+
+        for x in magazine:
+            hashmap[x] += 1
+
+        for x in ransomNote:
+            value = hashmap.get(x)
+            if value is None or value == 0:
+                return False
+            else:
+                hashmap[x] -= 1
+
+        return True
+```
+
+Python写法三：
+
+```python
 class Solution(object):
     def canConstruct(self, ransomNote, magazine):
         """
@@ -166,6 +210,7 @@ class Solution(object):
 ```
 
 Go：
+
 ```go
 func canConstruct(ransomNote string, magazine string) bool {
     record := make([]int, 26)

problems/0454.四数相加II.md

@@ -120,7 +120,7 @@ class Solution {
 
 Python：
 
-```
+```python
 class Solution(object):
     def fourSumCount(self, nums1, nums2, nums3, nums4):
         """
@@ -147,7 +147,31 @@ class Solution(object):
                 if key in hashmap:
                     count += hashmap[key]
         return count
+ 
+# 下面这个写法更为简洁，但是表达的是同样的算法
+# class Solution:
+#     def fourSumCount(self, nums1: List[int], nums2: List[int], nums3: List[int], nums4: List[int]) -> int:
+#         from collections import defaultdict
+
+#         hashmap = defaultdict(int)
+
+#         for x1 in nums1:
+#             for x2 in nums2:
+#                 hashmap[x1+x2] += 1
+        
+#         count=0
+#         for x3 in nums3:
+#             for x4 in nums4:
+#                 key = -x3-x4
+#                 value = hashmap.get(key)
+
+                  # dict的get方法会返回None（key不存在）或者key对应的value
+                  # 所以如果value==0，就会继续执行or，count+0，否则就会直接加value
+                  # 这样就不用去写if判断了
+            
+#                 count += value or 0
                 
+#         return count
 
 ```
 

problems/0707.设计链表.md

@@ -340,7 +340,7 @@ class MyLinkedList {
 ```
 
 Python：
-```python3
+```python
 # 单链表
 class Node: