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

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本文深入解析了TwoSumII算法，该算法针对已排序数组快速查找两个数使其和等于目标值的问题。通过双指针技巧，算法实现了高效查找，避免了重复元素的使用，确保了唯一解的准确性。

Two Sum II - Input array is sorted (E)

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.

题意

给定一个升序数组，找到数组中的两个数，使它们的和正好为目标值，返回这两个数的下标。

思路

Two Pointers。

代码实现

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

        return null;
    }
}