167. Two Sum II - Input Array Is Sorted

原创于 2022-03-03 22:26:10 发布 · 163 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

Leetcode 专栏收录该内容

24 篇文章

订阅专栏

本文介绍如何使用O(1)空间复杂度解决已排序数组中寻找两个数之和等于目标值的问题。通过迭代左右指针的方法，找到符合条件的两个数的下标并返回。实例代码演示了如何在Solution类中实现这个功能，适用于面试和技术挑战。

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 <= numbers.length.

Return the indices of the two numbers, index1 and index2, added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.

Example 1:

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. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

class Solution {
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
        int left=0,right=numbers.size()-1;
        vector<int> ret(2,-1);
        while (left<right){
            if (numbers[left] + numbers[right] > target)
                right --;
            else if (numbers[left] + numbers[right] < target)
                left ++;
            else{
                ret[0] = left+1;
                ret[1] = right+1;
                return ret;
                } 
               
        }
        return ret;
    }
};