Lintcode: Permutation Index II

Given a permutation which may contain repeated numbers, find its index in all the permutations of these numbers, which are ordered in lexicographical order. The index begins at 1.

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Example
Given the permutation [1, 4, 2, 2], return 3.

這裏須要考慮重複元素，有無重複元素最大的區別在於原來的1!, 2!, 3!...等須要除以重複元素個數的階乘。記錄重複元素個數一樣須要動態更新，引入哈希表這個萬能的工具較爲方便。

按照從數字低位到高位進行計算

 1 public class Solution {
 2     /**
 3      * @param A an integer array
 4      * @return a long integer
 5      */
 6     public long permutationIndexII(int[] A) {
 7         // Write your code here
 8         if (A==null || A.length==0) return new Long(0);
 9         int len = A.length;
10         HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
11         long index = 0, fact = 1, mulFact = 1;
12         for (int i=len-1; i>=0; i--) {
13             if (!map.containsKey(A[i])) {
14                 map.put(A[i], 1);
15             }
16             else {
17                 map.put(A[i], map.get(A[i])+1);
18                 mulFact *= map.get(A[i]);
19             }
20             int count = 0;
21             for (int j=i+1; j<len; j++) {
22                 if (A[i] > A[j]) count++;
23             }
24             index += count*fact/mulFact;
25             fact *= (len-i);
26         }
27         index = index + 1;
28         return index;
29     }
30 }