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

题目地址

https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/

题目描述

这是leetcode头号题目two sum的第二个版本，难度简单。

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.

前置知识

• 双指针

公司

• amazon

思路

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）

关键点解析

• 由于是有序的，因此双指针更好

代码

• 语言支持：JS，C++，Java，Python

Javascript Code:

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    const visited = {} // 记录出现的数字， 空间复杂度N

    for (let index = 0; index < numbers.length; index++) {
        const element = numbers[index];
        if (visited[target - element] !== void 0) {
            return [visited[target - element], index + 1]
        }
        visited[element] = index + 1;
    }
    return [];
};

C++ Code:

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int n = numbers.size();
        int left = 0;
        int right = n-1;
        while(left <= right)
        {
            if(numbers[left] + numbers[right] == target)
            {
                return {left + 1, right + 1};
            }
            else if (numbers[left] + numbers[right] > target)
            {
                right--;
            }
            else
            {
                left++;
            }
        }
        return {-1, -1};
    }
};

Java Code:

class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int n = numbers.length;
        int left = 0;
        int right = n-1;
        while(left <= right)
        {
            if(numbers[left] + numbers[right] == target)
            {
                return new int[]{left + 1, right + 1};
            }
            else if (numbers[left] + numbers[right] > target)
            {
                right--;
            }
            else
            {
                left++;
            }
        }
        
        return new int[]{-1, -1};
    }
}

Python Code:

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]

复杂度分析

• 时间复杂度：$O(N)$
• 空间复杂度：$O(1)$

更多题解可以访问我的LeetCode题解仓库：https://github.com/azl397985856/leetcode 。 目前已经30K star啦。

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