Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

翻译

给定一个整形数组和一个整数target，返回2个元素的下标，它们满足相加的和为target。

你可以假设每种输入只会对应一个答案,但是，数组中同一个元素不能使用两遍。

Kotlin实现1

fun twoSum(nums: IntArray, target: Int): IntArray? {
    for (i in nums.indices) {
        for (j in i + 1 until nums.size) {
            if (nums[j] + nums[i] == target) {
                return intArrayOf(i, j)
            }
        }
    }
    return intArrayOf(0, 0)
}

Kotlin实现2

fun twoSum(nums: IntArray, target: Int): IntArray? {
    val map: MutableMap<Int, Int> = HashMap()
    for (i in nums.indices) {
        val key = target - nums[i]
        if (map.containsKey(key)) {
            return intArrayOf(map[key]!!, i)
        }
        map[nums[i]] = i
    }
    return intArrayOf(0, 0)
}

参考：
https://leetcode.com/problems/two-sum/