lintcode-easy-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.

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

思路：

统计最高位之后最高位数小的个数，(n-1)!乘以这个数，就得到最高位比现在的数最高位小的时候的所有组合数。再把最高位去掉，重复这个过程，就得到了结果。

public class Solution {
/**
* @param A an integer array
* @return a long integer
*/
public long permutationIndex(int[] A) {
// Write your code here
if(A == null || A.length == 0)
return (long) 0;

long result = 0;

for(int i = 0; i < A.length; i++){
long num = 0;
for(int j = i + 1; j < A.length; j++){
if(A[j] < A[i])
num++;
}

result += num * fac(A.length - i - 1);
}

return result + 1;
}

public long fac(int num){
if(num == 0 || num == 1)
return 1;

long result = 1;

for(int i = 1; i <= num; i++)
result *= i;

return result;
}
}