# LeetCode 1. Two Sum
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.

You can return the answer in any order.

### Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

### Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

### Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

### Constraints:

* 2 <= nums.length <= 104
* -109 <= nums[i] <= 109
* -109 <= target <= 109
* Only one valid answer exists.

> Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

## 解題

##### Kotlin 1
雙迴圈的時間複雜度是O(n^2)
這一個跑分
Runtime: 244 ms, faster than 48.10% of Kotlin online submissions for Two Sum.
Memory Usage: 36.8 MB, less than 86.13% of Kotlin online submissions for Two Sum.

  class Solution {
        fun twoSum(nums: IntArray, target: Int): IntArray {
            for (i in nums.indices) {
                for (j in nums.indices) {
                    if (nums[i] + nums[j] == target && i != j) {
                        return intArrayOf(i, j)
                    }
                }
            }
            return intArrayOf()
        }
    }

網路上查用 map 比較快。試試。

###### tags: LeetCode