126 lines
3.5 KiB
Markdown
126 lines
3.5 KiB
Markdown
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## 题目地址
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https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/
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## 题目描述
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这是leetcode头号题目`two sum`的第二个版本，难度简单。
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```
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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.
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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.
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Note:
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Your returned answers (both index1 and index2) are not zero-based.
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You may assume that each input would have exactly one solution and you may not use the same element twice.
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Example:
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Input: numbers = [2,7,11,15], target = 9
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Output: [1,2]
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Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
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```
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## 思路
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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。
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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，
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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。
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> 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）
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## 关键点解析
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无
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## 代码
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* 语言支持：JS，Python
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Javascript Code:
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```js
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/*
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* @lc app=leetcode id=167 lang=javascript
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*
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* [167] Two Sum II - Input array is sorted
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*
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* https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/
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*
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* algorithms
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* Easy (49.46%)
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* Total Accepted: 221.8K
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* Total Submissions: 447K
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* Testcase Example: '[2,7,11,15]\n9'
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*
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* Given an array of integers that is already sorted in ascending order, find
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* two numbers such that they add up to a specific target number.
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*
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* The function twoSum should return indices of the two numbers such that they
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* add up to the target, where index1 must be less than index2.
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*
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* Note:
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*
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*
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* Your returned answers (both index1 and index2) are not zero-based.
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* You may assume that each input would have exactly one solution and you may
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* not use the same element twice.
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*
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*
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* Example:
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*
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*
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* Input: numbers = [2,7,11,15], target = 9
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* Output: [1,2]
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* Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.
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*
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*/
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/**
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* @param {number[]} numbers
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* @param {number} target
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* @return {number[]}
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*/
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var twoSum = function(numbers, target) {
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const visited = {} // 记录出现的数字， 空间复杂度N
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for (let index = 0; index < numbers.length; index++) {
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const element = numbers[index];
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if (visited[target - element] !== void 0) {
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return [visited[target - element], index + 1]
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}
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visited[element] = index + 1;
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}
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return [];
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};
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```
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Python Code:
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```python
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class Solution:
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def twoSum(self, numbers: List[int], target: int) -> List[int]:
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visited = {}
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for index, number in enumerate(numbers):
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if target - number in visited:
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return [visited[target-number], index+1]
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else:
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visited[number] = index + 1
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# 双指针思路实现
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class Solution:
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def twoSum(self, numbers: List[int], target: int) -> List[int]:
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left, right = 0, len(numbers) - 1
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while left < right:
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if numbers[left] + numbers[right] < target:
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left += 1
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if numbers[left] + numbers[right] > target:
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right -= 1
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if numbers[left] + numbers[right] == target:
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return [left+1, right+1]
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```
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