LeetCode 167. Two Sum II - Input array is sorted (Easy)

Description:

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

• Your returned answers (both index1 and index2) are not zero-based.
• You may assume that each input would have exactly one solution and you may not use the same element twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

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Analysis:

We can utilize two pointers to solve this problem.
We set two indexes to represent two pointers, at the beginning, one index points to the first element of the array and another index points to the last element of the array. Compare the sum of two elements that the two indexes point to. If $sum = target$, then the sum is the exactly unique solution. If $sum < target$, increase the smaller index by 1, and if $sum > target$, decrease the larger index by 1. Repeat the comparison and move until the solution is found.

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Code:

class Solution {
    public int[] twoSum(int[] numbers, int target) {
       int start = 0, end = numbers.length-1;
       int rs[] = new int[2];
       while(start < end) {
           int sum = numbers[start] + numbers[end];
           if(sum == target) {
               rs[0] = (start+1);
               rs[1] = (end+1);
               break;
           }else if(sum < target){
               start++;
           }else{
               end--;
           }
       }
        return rs;
    }
}

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Complexity:

Time complexity: $O(n)$
Space complexity: $O(1)$

来源：CSDN

作者：Boston_Kimisong

链接：https://blog.csdn.net/weixin_45744426/article/details/103752413