标签：
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.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.
思路：
吃惊的是LeetCode上O(N2)的最朴素的解法是可以AC的，个人选择了基于HashMap的O(N)的解法，先将数组存入后再遍历数组，寻找map中是否存在对应差值。
解法：
1 import java.util.Map; 2 3 public class Solution 4 { 5 public int[] twoSum(int[] nums, int target) 6 { 7 int[] returnArray = new int[2]; 8 Map<Integer, Integer> map = new HashMap<>(); 9 int subtractValue; 10 11 for(int i = 0; i < nums.length; i++) 12 map.put(nums[i], i); 13 for(int i = 0; i < nums.length; i++) 14 { 15 subtractValue = target - nums[i]; 16 if(map.containsKey(subtractValue) && map.get(subtractValue) != i) 17 { 18 returnArray[0] = map.get(subtractValue); 19 returnArray[1] = i; 20 break; 21 } 22 } 23 24 return returnArray; 25 } 26 }
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原文地址：http://www.cnblogs.com/wood-python/p/5834393.html
