* 197. Permutation Index【LintCode by java】

Description

Given a permutation which contains no repeated number, find its index in all the permutations of these numbers, which are ordered in lexicographical order. The index begins at 1.

Example

Given [1,2,4], return 1.

解题：至今仍然不理解这个“字典中的顺序”是什么意思，做个标记日后有时间再看。照着人家的代码敲得：https://www.cnblogs.com/theskulls/p/4881142.html      代码如下：

 public class Solution {
     /**
      * @param A: An array of integers
      * @return: A long integer
      */
     public long permutationIndex(int[] A) {
         // write your code here
         long index = 0;
         long position = 2;
         long factor = 1;
         for (int p = A.length - 2; p >= 0; p--) {
             long successors = 0;
             for (int q = p + 1; q < A.length; q++) {
                 if (A[p] > A[q]) {
                     successors++;
                 }
             }
         index += (successors * factor);
         factor *= position;
         position++;
         }
         index = index + 1;
         return index;
         }
 }