题目

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。

你可以假定每个输入，都会恰好有一个满足条件的返回结果。

Hints

Related Topics: Array, Hash Table

如果简单地想O(n^2)的算法很容易实现

但是可以利用Hash Table来实现O(n)的算法

代码

Java

class Solution {

    public int[] twoSum(int[] nums, int target) {

        Map<Integer,Integer> mymap = new HashMap<>();

        for(int i=0;i<nums.length;i++){

            Integer index = mymap.get(target-nums[i]);

            if(index==null){

                mymap.put(nums[i],i);

            }else{

                return new int[]{i,index};

            }

        }

        return new int[]{0,0};

    }

}

//solution from discuss

class Solution {

    public int[] twoSum(int[] numbers, int target) {

        int[] result = new int[2];

        Map<Integer, Integer> map = new HashMap<Integer, Integer>();

        for (int i = 0; i < numbers.length; i++) {

            if (map.containsKey(target - numbers[i])) {

                result[1] = i + 1;

                result[0] = map.get(target - numbers[i]);

                return result;

            }

            map.put(numbers[i], i + 1);

        }

        return result;

    }

}

Python

class Solution(object):

    def twoSum(self, nums, target):

        """

        :type nums: List[int]

        :type target: int

        :rtype: List[int]

        """

        mymap = {}

        for i in range(len(nums)):

            if nums[i] in mymap:

                return [mymap[nums[i]],i]

            else:

                mymap[target-nums[i]] = i

        return [0,0]