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.

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，代码比较简单， 不贴了。

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

无

代码

/*
 * @lc app=leetcode id=167 lang=javascript
 *
 * [167] Two Sum II - Input array is sorted
 *
 * https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/description/
 *
 * algorithms
 * Easy (49.46%)
 * Total Accepted:    221.8K
 * Total Submissions: 447K
 * Testcase Example:  '[2,7,11,15]\n9'
 *
 * 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
 * @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 [];
};