6

Best method to search for an array?

 3 years ago
source link: https://www.codesd.com/item/best-method-to-search-for-an-array.html
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
neoserver,ios ssh client

Best method to search for an array?

advertisements

I have an array (nodes[][]) that contains values of effective distances that looks something like this:

__                 __
|1    0.4  3         |
|0.4  1    0         |
|3    3.2  1   ...   |
|0.8  4    5         |
|0    0    1         |
--                  --

Where the first value, node[0][0] is the distance from node 0 to node 0 which is 1.
So the distance from node 2 to node 1 is 3.2 (node[2][1]=3.2)
I need, given a node column, to search through the rows to find the farthest distance, while not picking itself (node[1][1])
The method I was thinking to do something like this:

int n=0;
currentnode=0;  //this is the column I am searching now
if(currentnode==n)
   n++;
best=node[n][currentnode];
nextbest=node[n++][currentnode];
if(nextbest>best)
   best=nextbest;
else
  for(int x=n;x<max;x++)    //max is the last column
  {
   if(currentnode==n)
      continue;
   nextbest=node[x][currentnode];
   if(nextbest>best)
      best=nextbest;
  }

I can't think of a better method to do this. I could use functions to make it shorter but this is GENERALLY what I am thinking about using. After this I have to loops this to go to the next column that the best distance returns and do this routine again.

You can simplify it quite a bit. A lot of your checks and temporary variables are redundant. Here's a small function that performs your search. I've renamed most of the variables to be a little more precise what their roles are.

int maxDistance(int fromNode) {
    int max = -1;

    for (int toNode = 0; toNode < nodeCount; ++toNode)
    {
        if (fromNode != toNode && nodes[toNode][fromNode] > max) {
            max = node[toNode][fromNode];
        }
    }

    return max;
}

report

Recommend

About Joyk


Aggregate valuable and interesting links.
Joyk means Joy of geeK