【leetcode系列】167. 两数之和 II - 输入有序数组

给定一个已按照升序排列 的有序数组，找到两个数使得它们相加之和等于目标数。

函数应该返回这两个下标值 index1 和 index2，其中 index1 必须小于 index2。

说明:

返回的下标值（index1 和 index2）不是从零开始的。
你可以假设每个输入只对应唯一的答案，而且你不可以重复使用相同的元素。
示例:

输入: numbers = [2, 7, 11, 15], target = 9
输出: [1,2]
解释: 2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted
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/*
 * @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 [];
};