leetcode 167. Two Sum II

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

/*思路：

首先先定义两个指针分别指向该vector的头和尾

1.如果numbers[left] + numbers[right] <  target; 则左指针向右移到一个较大的数

2.如果numbers[left] + numbers[right] > target; 则右指针向左移到一个较小的数

3.如果numbers[left] + numbers[right] = target; 则返回  left + 1 和 right + 1下标   （注，题目中是以一为下标开始）

/*

// 具体函数的定义如下：

#include <iostream>

#include <vector>

using namespace std;

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int> set(2, -1);
        int left = 0;
        int right = numbers.size() - 1;
        while(left < right)
        {
            long long int sum = numbers[left] + numbers[right]; //防止两个 int 相加 溢出int 
            if(sum == target)
            {
                set[0] = left + 1;
                set[1] = right + 1;
                return set;
            }
            else if(sum < target) left ++; //左指针向右移 
            else right --;  //右指针向左移 
        }
    }
};