167. Two Sum II

167. Two Sum II - Input array is sorted

题目

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. Please note that 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.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

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翻译

给定一个已经按升序排序的整数数组，找到两个数字，使它们相加到一个特定的目标数。

函数twoSum应该返回两个数字的索引，使它们相加到目标，其中index1必须小于index2。请注意，您返回的答案（index1和index2）都不是基于零的。

你可以假设每个输入将有恰好一个解决方案，你可能不会使用相同元素的两倍。

输入： numbers = { 2，7，11，15 }，target = 9

输出： index1 = 1，index2 = 2

解题思路

典型的从数组的两侧开始遍历，进行求和判断不断缩小范围，时间复杂度为O(n)

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

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