Rampage Double Tube Rear Bumper for 1976-1983 Jeep CJ5

Rampage Double Tube Rear Bumper for 1976-1983 Jeep CJ5

Engineered for off-road enthusiasts and classic Jeep restorers, the Rampage Double Tube Rear Bumper delivers rugged durability and a commanding presence. This bumper is robotically welded to exacting specifications, ensuring long-lasting performance and resistance to the harshest environments. The black dual powder coat finish is applied over a dense zinc undercoating, providing superior protection against rust, flaking, and peeling.

• One-piece, robot# 1. 两数之和（简单）[数组，哈希表（HashMap）] ## 题目描述及示例 ### 题目描述 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。 你可以按任意顺序返回答案。 **示例 1：** ``` 输入：nums = [2,7,11,15], target = 9 输出：[0,1] 解释：因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。 ``` **示例 2：** ``` 输入：nums = [3,2,4], target = 6 输出：[1,2] ``` **示例 3：** ``` 输入：nums = [3,3], target = 6 输出：[0,1] ``` **提示：** * `2 <= nums.length <= 104` * `-109 <= nums[i] <= 109` * `-109 <= target <= 109` * **只会存在一个有效答案** **进阶：**你可以想出一个时间复杂度小于 `O(n2)` 的算法吗？ --- ### 我的回答 ```java class Solution { public int[] twoSum(int[] nums, int target) { Map map = new HashMap<>(); for (int i = 0; i < nums.length; i++) { int complement = target - nums[i]; if (map.containsKey(complement)) { return new int[]{map.get(complement), i}; } map.put(nums[i], i); } throw new IllegalArgumentException(