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：
- 利用hash map
- 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 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}; }}