167. Sum of two numbers II - enter an ordered array

Title Description

I'll give you a subscript from 1 The starting array of integers numbers , The array has been pressed Non decreasing order , Please find out from the array that the sum satisfying the addition is equal to the target number target Two numbers of . Let these two numbers be numbers[index1] and numbers[index2] , be 1 <= index1 < index2 <= numbers.length .

In length 2 Array of integers for [index1, index2] Returns the subscript of these two integers in the form of index1 and index2.

You can assume that each input Only corresponding to the only answer , And you Can not be Reuse the same elements .

The solution you design must use only constant level extra space .

Example 1：

Input ：numbers = [2,7,11,15], target = 9
Output ：[1,2]
explain ：2 And 7 The sum is equal to the number of targets 9 . therefore index1 = 1, index2 = 2 . return [1, 2] .
Example 2：

Input ：numbers = [2,3,4], target = 6
Output ：[1,3]
explain ：2 And 4 The sum is equal to the number of targets 6 . therefore index1 = 1, index2 = 3 . return [1, 3] .
Example 3：

Input ：numbers = [-1,0], target = -1
Output ：[1,2]
explain ：-1 And 0 The sum is equal to the number of targets -1 . therefore index1 = 1, index2 = 2 . return [1, 2] .

Tips ：

2 <= numbers.length <= 3 * 104
-1000 <= numbers[i] <= 1000
numbers Press Non decreasing order array
-1000 <= target <= 1000
There is only one valid answer

Answer key

Method 1 ： Double pointer

The optimal

class Solution {
    
public:
    vector<int> twoSum(vector<int>& numbers, int target) {
    
        
 // Double pointer   Time complexity O(n)
int i=0,j=numbers.size()-1;  
       while(j>i)
        {
    
            if(numbers[i]+numbers[j]>target){
    j--;}
            else if(numbers[i]+numbers[j]==target){
    return {
    i+1,j+1};}
            else i++;
        }

return {
    -1,-1};

Method 2 ： Two points search

// Two points search  
 for (int i = 0; i < numbers.size(); ++i) {
    
            int low = i + 1, high = numbers.size() - 1,k = target - numbers[i];
            while (low <= high)
            {
    
                int mid = low + (high - low)/2 ;
                if (numbers[mid] == k) 
                {
    
                    return {
    i + 1, mid + 1};
                } 
                else if (numbers[mid] > k) 
                {
    
                    high = mid - 1;
                } 
                else 
                {
    
                    low = mid + 1;
                }
            }
        }
        return {
    -1, -1};
    }


// This is stupid , Use the one above 
int i = 0,k=0;
int left=0 , right=numbers.size(),mid=0;

while(left<=right)
{
      
    left=i+1;
    k = target-numbers[i];
    mid = left+(right-left)/2;
    if(numbers[mid]>k){
    right = mid-1;}
    else if(numbers[mid]==k)return {
    i+1,mid+1};
    else if(numbers[mid]<k)
    {
    
        for(mid = mid+1;mid<right;mid++)
        {
        if(numbers[mid]>k)break;
            else if(numbers[mid]==k)return {
    i+1,mid+1};
        }       
        ++i;  
    }
} 
return {
    0,0};
}
};