1. Two Sum

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].

难度：easy

题目：
给定一整数数组，返回使得数组中两元素的和为特定整数的下标。
假定每个输入都会有且仅有一个结果，不可以重复使用同一个元素。

思路：

1. 利用hash map
2. sort 后从两头向中间夹挤

Runtime: 352 ms, faster than 96.77% of Scala online submissions for Two Sum.

object Solution {
    def twoSum(nums: Array[Int], target: Int): Array[Int] = {
        import scala.collection.mutable.Map
        val mii: Map[Int, Int] = Map()
        for (i ← 0 until nums.size) {
            val adder = target - nums(i)
            if (mii.contains(adder)) {
                return Array(mii.get(adder).get, i)
            }
            mii += (nums(i) → i)
        }
        Array(0, 0)
    }
}

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> mii = new HashMap<>();
        /*
         * 从前向后遍历，既然是一定会有匹配的数，那只能是前面找不到对应的，后面可以找到找出的index应该是前的。所以后面遍历的i作为右边index 
         */
        for (int i = 0; i < nums.length; i++) {
            if (mii.containsKey(target - nums[i])) {
                return new int[] {mii.get(target - nums[i]), i};
            }
            mii.put(nums[i], i);
        }

        return new int[] {0, 0};
    }
}