@@ -28,7 +28,7 @@ Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
2828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。
2929
3030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,
31- 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 。
31+ 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 。
3232
3333> 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)
3434
@@ -40,6 +40,10 @@ Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
4040
4141## 代码
4242
43+ * 语言支持:JS,Python
44+ 45+ Javascript Code:
46+ 4347``` js
4448/*
4549 * @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.
5660 *
5761 * Given an array of integers that is already sorted in ascending order, find
5862 * two numbers such that they add up to a specific target number.
59- *
63+ *
6064 * The function twoSum should return indices of the two numbers such that they
6165 * add up to the target, where index1 must be less than index2.
62- *
66+ *
6367 * Note:
64- *
65- *
68+ *
69+ *
6670 * Your returned answers (both index1 and index2) are not zero-based.
6771 * You may assume that each input would have exactly one solution and you may
6872 * not use the same element twice.
69- *
70- *
73+ *
74+ *
7175 * Example:
72- *
73- *
76+ *
77+ *
7478 * Input: numbers = [2,7,11,15], target = 9
7579 * Output: [1,2]
7680 * Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
77- *
81+ *
7882 */
7983/**
8084 * @param {number[]} numbers
@@ -89,13 +93,33 @@ var twoSum = function(numbers, target) {
8993 if (visited[target - element] !== void 0 ) {
9094 return [visited[target - element], index + 1 ]
9195 }
92- visited[element] = index + 1 ;
96+ visited[element] = index + 1 ;
9397 }
9498 return [];
9599};
96- 97- 98100```
99101
100- 101- 102+ Python Code:
103+ 104+ ``` python
105+ class Solution :
106+ def twoSum (self numbers : List[int ], target : int ) -> List[int ]:
107+ visited = {}
108+ for index, number in enumerate (numbers):
109+ if target - number in visited:
110+ return [visited[target- number], index+ 1 ]
111+ else :
112+ visited[number] = index + 1
113+ 114+ # 双指针思路实现
115+ class Solution :
116+ def twoSum (self numbers : List[int ], target : int ) -> List[int ]:
117+ left, right = 0 , len (numbers) - 1
118+ while left < right:
119+ if numbers[left] + numbers[right] < target:
120+ left += 1
121+ if numbers[left] + numbers[right] > target:
122+ right -= 1
123+ if numbers[left] + numbers[right] == target:
124+ return [left+ 1 , right+ 1 ]
125+ ```
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