The beauty of programming: quickly finding two numbers that meet the conditions

Can we quickly find two numbers in an array so that the sum of these two numbers is equal to a given value? To simplify the process, we assume that there must be at least one set of compliant solutions in this array.

Method 1:

The most direct method is the exhaustive method. The complexity is O (N ^ 2 );

 

Method 2:

Use sum to subtract a [I] And then search for sum-a [I]. If the sum is in the array, the result is a search. You can use binary search; the sorting complexity is O (nlgn), the search complexity is O (lgn), and the final complexity is O (nlgn );

, You can also use hash to search, but the space complexity increases by O (N );

 

Method 3: sort the order first, and then, I = 0, j = n-1. Check whether arr [I] + arr [j] is equal to sum. If it is less than sum, I ++; greater than sum, j --; Complexity: O (N) + O (NlgN );

 
 

for(i=0;j=n-1;i<j;)    if(arr[i]+arr[j]==sum)         return (i,j);    else if(arr[i]+arr[j]<sum)        i++;    else        j--;    return (i,j); for(i=0;j=n-1;i<j;)   if(arr[i]+arr[j]==sum)     return (i,j);   else if(arr[i]+arr[j]<sum)    i++;   else    j--;   return (i,j);