1, and a certain time to find the number of two in the array

Title Requirements:

Given an array of integers and a target value, find the two numbers in the array and the target values.

You can assume that each input corresponds to only one answer, and that the same element cannot be reused.

Ideas:

Sum must be, to ask two numbers, you need two pointers, two pointers a large one small do not want to wait, first let the small pointer do not move, let the large pointer move backwards, know that appears greater than or equal to the need. Equal to the output result, the small pointer moves backward one bit, and a pointer that is larger than the time of the move appears greater than or equal to the required sum. And so on, the result is obtained. Equivalent to a reverse "median theorem".

Code:

 Public classOne { Public Static voidMain (string[] args) {int[] Nums = {1,2,3,4,5,6,7,8,9}; inttarget = 10;  for(inta=0;a<nums.length;a++)    {         for(intb=a+1;b<nums.length;b++)             if(Nums[a]+nums[b] = =target) System.out.println ("There has" +nums[a]+ "and" +nums[b]+ "=" +target); }}}

1, and a certain time to find the number of two in the array