LeetCode - 167.Two Sum II - Input array is sorted_双指针在有序数组中找出两个数,使他们的和为-优快云博客

本文链接：https://blog.youkuaiyun.com/Dongdong2333/article/details/105927168

本文介绍了一种在已排序的数组中寻找两个数使其和等于特定目标值的有效算法。通过双指针技术，该算法能在O(n)时间内找到解决方案，避免了重复使用相同元素的情况。

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概述

Two Sum II - Input array is sorted
Easy

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.

即在有序数组中找出两个数，使他们的和为target。并且两次不能使用相同的元素。

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.

思路

使用双指针，一个指针指向较小元素(number[0])，另一个指向较大元素(number[numbers.length-1])。
分别从两边开始遍历。
如果sum==target，返回下标。
sum>target，令sum变小，即j–；
sum<target，令sum变大，即i++；
否则，返回null。

代码

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