feat: 167.two-sum-ii-input-array-is-sorted add Python3 implement… (az…

…l397985856#101)

* feat: 167.two-sum-ii-input-array-is-sorted add Python3 implementation

* Update 167.two-sum-ii-input-array-is-sorted.md

problems/167.two-sum-ii-input-array-is-sorted.md

@@ -28,7 +28,7 @@ Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
 由于题目没有对空间复杂度有求，用一个hashmap  存储已经访问过的数字即可。
 
 假如题目空间复杂度有要求，由于数组是有序的，只需要双指针即可。一个left指针，一个right指针，
-如果left + right 值 大于target 则 right左移动， 否则left右移，代码比较简单， 不贴了。
+如果left + right 值 大于target 则 right左移动， 否则left右移，代码见下方python code。
 
 > 如果数组无序，需要先排序（从这里也可以看出排序是多么重要的操作）
 
@@ -40,6 +40,10 @@ Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
 
 ## 代码
 
+* 语言支持：JS，Python
+
+Javascript Code:
+
 ```js
 /*
  * @lc app=leetcode id=167 lang=javascript
@@ -56,25 +60,25 @@ Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
  *
  * Given an array of integers that is already sorted in ascending order, find
  * two numbers such that they add up to a specific target number.
- * 
+ *
  * The function twoSum should return indices of the two numbers such that they
  * add up to the target, where index1 must be less than index2.
- * 
+ *
  * Note:
- * 
- * 
+ *
+ *
  * Your returned answers (both index1 and index2) are not zero-based.
  * You may assume that each input would have exactly one solution and you may
  * not use the same element twice.
- * 
- * 
+ *
+ *
  * Example:
- * 
- * 
+ *
+ *
  * Input: numbers = [2,7,11,15], target = 9
  * Output: [1,2]
  * Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
- * 
+ *
  */
 /**
  * @param {number[]} numbers
@@ -89,13 +93,33 @@ var twoSum = function(numbers, target) {
         if (visited[target - element] !== void 0) {
             return [visited[target - element], index + 1]
         }
-        visited[element] = index + 1;   
+        visited[element] = index + 1;
     }
     return [];
 };
-
-
 ```
 
-
-
+Python Code:
+
+```python
+class Solution:
+    def twoSum(self, numbers: List[int], target: int) -> List[int]:
+        visited = {}
+        for index, number in enumerate(numbers):
+            if target - number in visited:
+                return [visited[target-number], index+1]
+            else:
+                visited[number] = index + 1
+
+# 双指针思路实现
+class Solution:
+    def twoSum(self, numbers: List[int], target: int) -> List[int]:
+        left, right = 0, len(numbers) - 1
+        while left < right:
+            if numbers[left] + numbers[right] < target:
+                left += 1
+            if numbers[left] + numbers[right] > target:
+                right -= 1
+            if numbers[left] + numbers[right] == target:
+                return [left+1, right+1]
+```