Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
\n\nYou can return the answer in any order.
\n\n\n
Example 1:
\n\n\nInput: nums = [2,7,11,15], target = 9\nOutput: [0,1]\nExplanation: Because nums[0] + nums[1] == 9, we return [0, 1].\n\n\n
Example 2:
\n\n\nInput: nums = [3,2,4], target = 6\nOutput: [1,2]\n\n\n
Example 3:
\n\n\nInput: nums = [3,3], target = 6\nOutput: [0,1]\n\n\n
\n
Constraints:
\n\n2 <= nums.length <= 104-109 <= nums[i] <= 109-109 <= target <= 109\nFollow-up: Can you come up with an algorithm that is less than
O(n2) time complexity?",
"translatedTitle": "两数之和",
"translatedContent": "给定一个整数数组 nums 和一个整数目标值 target，请你在该数组中找出 和为目标值 target 的那 两个 整数，并返回它们的数组下标。
你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。
\n\n你可以按任意顺序返回答案。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [2,7,11,15], target = 9\n输出：[0,1]\n解释：因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。\n\n\n
示例 2：
\n\n\n输入：nums = [3,2,4], target = 6\n输出：[1,2]\n\n\n
示例 3：
\n\n\n输入：nums = [3,3], target = 6\n输出：[0,1]\n\n\n
\n\n
提示：
\n\n2 <= nums.length <= 104-109 <= nums[i] <= 109-109 <= target <= 109\n\n
进阶：你可以想出一个时间复杂度小于 O(n2) 的算法吗？
x, we have to scan the entire array to find the next number y which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?",
"The second train of thought is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?"
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"sampleTestCase": "[2,7,11,15]\n9",
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