Sword finger offer notes: t57 - I. and two numbers of S

T57 - I. And for s Two numbers of

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

solution 1

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int left =0, right = nums.length-1;
        int[] two = new int[2];
        while(left<=right){
            if(target-nums[right]<=0){
                right--;
            }else{
                if(target==nums[left]+nums[right]){
                    two[0]=nums[left];
                    two[1]=nums[right];
                }else if(target>nums[left]+nums[right]){
                    left++;
                }else{
                    return two;
                }
            }
        }
        return two;
    }
}

It's overtime

solution 2

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int left =0, right = nums.length-1;
        while(left<=right){
            int sum =nums[left]+nums[right];
            if(sum>target){
                right--;
            }else if(sum<target){
                left++;
            }else{
                return new int[]{nums[left],nums[right]};
            }
        }
       return new int[0]; 
    }
}

Execution time ：2 ms, In all  Java  Defeated in submission **95.08%** Users of

Memory consumption ：55.1 MB, In all  Java  Defeated in submission **86.87%** Users of

Time complexity ：O(n), Spatial complexity ：O(1)