Two Sum II - Input array is sorted---LeetCode167

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题目描述：
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

解释

由于题目给出的是从小到大排序的数组，并且对于每个给定的target，都会有确定的index值。这里是一种时间复杂度为O（n）的解法。
使用两个指针left和right，left指向数组的起始位置，right指向数组的末尾。用temp记录numbers[left]和numbers[right]的和，若temp==target，返回left+1和right+1；若temp值小于target，则left右移一位；反之，right左移一位，直到找到目标index值。

代码块

public int[] twoSum(int[] numbers,int target){
        if (numbers==null || numbers.length<1)
            return null;
        int index[]= new int[2];
        int left=0;
        int right=numbers.length-1;
        while(left<right){
            int temp=numbers[left]+numbers[right];
            if(temp==target){
                index[0]=left+1;
                index[1]=right+1;
                return index;
            }else if(temp<target){
                left++;
            }else {
                right--;
            }
        }
        return null;
    }