Description

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.

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Example:

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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测试用例

输入：3,3 6
输出： 0,1

输入：2,3,3 5
输出： 0,1
下面代码没有通过这个测试，但是A了。题目说结果是确定的，应该就是不考虑这种情况。

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思路

1. 先排序，最快O(nlogn)。再左右夹逼。比较暴力。
2. 利用hash表，时间复杂度为O(n)。下面代码是基于第二种方法。

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代码

vector<int> twoSum(vector<int>& nums, int target) {
       map<int,int> hashMap;
	for(int i = 0; i<nums.size(); i ++) hashMap[nums[i]] = i;
	
	//cout<<hashMap.size()<<endl;
	
	vector<int> ans;
	for(int i = 0; i<nums.size(); i ++){
		const int gap = target - nums[i];
		if(hashMap[gap] > 0){
			ans.push_back(i);
			ans.push_back(hashMap[gap]);
			break;
		}
	}
	return ans;
   }

官方java答案

public int[] twoSum(int[] nums, int target) {
   Map<Integer, Integer> map = new HashMap<>();
   for (int i = 0; i < nums.length; i++) {
       int complement = target - nums[i];
       if (map.containsKey(complement)) {
           return new int[] { map.get(complement), i };
       }
       map.put(nums[i], i);
   }
   throw new IllegalArgumentException("No two sum solution");
  }