* 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;
}
}
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