And is two numbers of S - two questions per day

The finger of the sword Offer 57. And for s Two numbers of

It's time to get up and brush questions every morning , It's really interesting to brush questions , Two questions a day , Come on ️

Enter an ascending array and a number s, Find two numbers in an array , So that their sum is exactly s. If the sum of many pairs of numbers is equal to s, Then output any pair .

Example 1：

Input ：nums = [2,7,11,15], target = 9
Output ：[2,7] perhaps [7,2]
Example 2：

Input ：nums = [10,26,30,31,47,60], target = 40
Output ：[10,30] perhaps [30,10]

Limit ：

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^6

Ideas

Algorithm ： Double pointer .

First, because ordered arrays , So the basic judgment is double pointer , Add the elements of the array corresponding to the left and right pointers ,

1. Equal to the target array , Assign to array , return , Remember to return
2. Larger than the target array , The right pointer moves to the left
3. Smaller than target array , The left pointer moves to the left

        int[] res = new int[2];
        int left = 0;
        int right = nums.length - 1;
        int sum = 0;
        while (left < right) {
    
            sum = nums[left] + nums[right];
            if (sum == target) {
    
                  res[0]=nums[left];
                  res[1]=nums[right];
                  return res;
            } else if (sum > target) {
    
                right--;
            } else {
    
                left++;
            }
        }
        return res;