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].

  /**
     * 暴力破解法
     * @param arr
     * @param target
     * @return
     */
    public static int[] TwoSum1(int[] arr,int target)
    {
        for(int i = 0; i < arr.length;i++)
        {
            int temp = target - arr[i];
            for(int j=i+1; j < arr.length; j++)
            {
                if(temp == arr[j])
                {
                    return new int[]{i,j};
                }
            }
        }
        return null;
    }

    /**
     * 利用map先将数组中的值放入
     * @param arr
     * @param target
     * @return
     */
    public static int[] TwoSum2(int[] arr, int target)
    {
        Map<Integer,Integer> map = new HashMap<>();
        for(int i = 0 ; i < arr.length; i++)
        {
            map.put(arr[i],i);
        }
        for(int j = 0; j < arr.length; j++)
        {
            int temp = target-arr[j];
            if(map.containsKey(temp)) {
                return new int[]{j, map.get(temp)};
            }
        }
        return null;
    }

    /**
     * 先不把arr中的元素放入map,逆向思维
     * @param arr
     * @param target
     * @return
     */
    public static int[] TwoSum3(int[] arr,int target)
    {
        Map<Integer,Integer> map = new HashMap<>();
        for(int i = 0; i < arr.length;i++)
        {
            int temp = target -arr[i];
            if(map.containsKey(temp))
            {
                return new  int[]{map.get(temp),i};
            }
            map.put(arr[i],i);
        }
        return null;
    }