[Cocos2dx] an algorithm for making the landlords & amp; quot; sort

When building a landlord, there is another sort of cards after the cards are issued, which is one of the issues to consider. The sorting here is not a general simple bubble sort, my idea is to arrange the card 0-51, with 13 as a benchmark classification, for example, 0-12 respectively correspond to the peach A-K, 13-25 corresponds to the peach A-K, 26-38 corresponds to the square A-K, 39-51 is plum blossom A-K. The server then sorts the 17 numbers allocated to each player into card arrays. The algorithm is given below. Thanks to @!

Code:

Using System; using System. collections. generic; using System. linq; using System. text; namespace random sorting {class Program {static void Main (string [] args) {int [] array = {1, 2, 6, 33, 44, 23, 36, 5, 9, 14,0, 14,15}; // shuffling from large to small // int [] array = {1, 2, 3, 4, 5, 6, 0 }; bubbleSortFunction (array); foreach (int I in array) {Console. write (I + ",") ;}// bubble sort (traversal starts from the starting position of the array, with a large number of benchmarks: sinking a large number) private static void Bubb LeSortFunction (int [] array) {int length = array. length; int temp; bool hasExchangeAction; // record whether two adjacent numbers in this large loop have been swapped (if not SWAps, the array is already ordered) for (int I = 0; I <length-1; I ++) // There are N numbers in the array, then you can use the N-1 to complete the Large Loop {hasExchangeAction = false; // each large loop assumes that the array is ordered for (int j = 0; j <length-I-1; j ++) // traverses from the array subscript 0, (length-I-1 is the calculated large number.) {int p1 = array [j] % 13; int p2 = array [j + 1] % 13; if (p1 = 0) p1 = 14; if (P2 = 0) p2 = 14; if (p1 = 1) p1 = 15; if (p2 = 1) p2 = 15; if (array [j] = 52) p1 = 16; if (array [j + 1] = 52) p2 = 16; if (array [j] = 53) p1 = 17; if (array [j + 1] = 53) p2 = 17; if (p1 <p2) // compare two adjacent numbers, if the preceding number is greater than the following number, the two adjacent numbers are swapped. {temp = array [j]; array [j] = array [j + 1]; array [j + 1] = temp; hasExchangeAction = true; // swapping occurred} if (! HasExchangeAction) // if no swaps have occurred, the array is already ordered and the loop {break ;}}} exists ;}}}}}

Result:

Note: according to the rules I mentioned above, we will find that the cards are actually sorted from big to small. The leftmost is Hongtao 2, then taotao 2, and then Hongtao, and so on!