167. Two Sum II



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 thesame element twice.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

解法一：双指针，一个从左边开始，一个从右边开始，如果和比target小，则左边指针右移，如果和比target大，则右边指针左移，直到和等于target为止

public int[] twoSum(int[] numbers, int target) {        int left = 0, right = numbers.length - 1;int[] res = new int[2];while (left != right) {while (target > numbers[left] + numbers[right])left++;while (target < numbers[left] + numbers[right])right--;if (target == numbers[left] + numbers[right]) {res[0] = left + 1;res[1] = right + 1;return res;};}return res;    }

//解法二：很简单的思路但是不符合题目may not use the same element twice的要求

public class Solution {    public int[] twoSum(int[] numbers, int target) {        int[] ret=new int[2];        for(int i=0;i<numbers.length;i++)        {            int a=target - numbers[i];            for(int j=i+1;j<numbers.length;j++)            {                if(numbers[j]==a)                {                    ret[0] = i + 1;                    ret[1] = j + 1;                    return ret;                }            }        }        return ret;            }}