167 2Sum II - input array is sorted

来源：Leetcode

原帖：https://leetcode.com/problems/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.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
Solution: Two Pointers Binary Search

思考：
如果数组是顺序的，可以使用two pointers或者binary search去搜索。binary search更快。

代码：
1] two pointers.

 class Solution {  
 public:  
     vector<int> twoSum(vector<int> &numbers, int target) {  
         int l = 0, r = numbers.size()-1;  
         while (l < r) {  
             int sum = numbers[l] + numbers[r];  
             if (sum == target) break;  
             else if (sum < target) l++;  
             else r--;  
         }  
         return {l + 1, r + 1};  
     }  
 };  

2] binary search

 class Solution {  
 public:  
     vector<int> twoSum(vector<int> &numbers, int target) {  
         for (int i = 0; i < numbers.size()-1; ++i) {  
             int idx = binSearch(numbers, target - numbers[i], i);  
             if (idx != -1) {  
                 return {i+1, idx+1};  
             }  
         }    
         return {-1, -1};  
     }  
   
     int binSearch(vector<int> &numbers, int target, int start) {  
         int l = start + 1, r = numbers.size() - 1;  
         while (l <= r) {  
             int m = (l + r) / 2;  
             if (target == numbers[m]) return m;  
             else if (target < numbers[m]) r = m - 1;  
             else l = m + 1;  
         }  
         return -1;  
     }  
 };