add permutation index

Author: billryan

SUMMARY.md

@@ -113,6 +113,7 @@
    * [Next Permutation](exhaustive_search/next_permutation.md)
    * [Previous Permuation](exhaustive_search/previous_permuation.md)
    * [Unique Binary Search Trees II](exhaustive_search/unique_binary_search_trees_ii.md)
+   * [Permutation Index](exhaustive_search/permutation_index.md)
 * [Dynamic Programming - 动态规划](dynamic_programming/README.md)
    * [Triangle](dynamic_programming/triangle.md)
    * [Knapsack - 背包问题](dynamic_programming/knapsack.md)

exhaustive_search/permutation_index.md

@@ -0,0 +1,50 @@
+# Permutation Index
+
+## Source
+
+- lintcode: [(197) Permutation Index](http://www.lintcode.com/en/problem/permutation-index/)
+
+```
+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.
+```
+
+## 题解1 - 双重 for 循环
+
+### Java
+
+```java
+public class Solution {
+    /**
+     * @param A an integer array
+     * @return a long integer
+     */
+    public long permutationIndex(int[] A) {
+        if (A == null || A.length == 0) return 0;
+
+        long index = 1;
+        long factor = 1;
+        for (int i = A.length - 1; i >= 0; i--) {
+            int rank = 0;
+            for (int j = i + 1; j < A.length; j++) {
+                if (A[i] > A[j]) rank++;
+            }
+            index += rank * factor;
+            factor *= (A.length - i);
+        }
+
+        return index;
+    }
+}
+```
+
+## 题解2 - 哈希表
+
+## Reference
+
+- [Permutation Index](http://www.geekviewpoint.com/java/numbers/permutation_index)
+- [permutation-index | 九章算法](http://www.jiuzhang.com/solutions/permutation-index/)