Friday, April 15, 2016

LintCode - Permutation Index


https://segmentfault.com/a/1190000004683277
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.
http://algorithm.yuanbin.me/zh-hans/exhaustive_search/permutation_index.html
例如[4,2,1],对于4, 比4小的有2和1, 在4这一位,对应的factor为2! (factor的含义是在当前位置数值指定的情况下,变化index大于当前数值的数字,所可能的到的不同排列的个数), 因此在index=0时,可能的排列一共有2*2!. 同理,对于[4,2,1],结果为2*2! + 1*1 + 0*1 + 1 = 6

以序列1, 2, 4为例,其不同的排列共有 3!=6 种,以排列[2, 4, 1]为例,若将1置于排列的第一位,后面的排列则有 2!=2 种。将2置于排列的第一位,由于[2, 4, 1]的第二位4在1, 2, 4中为第3大数,故第二位可置1或者2,那么相应的排列共有 2 * 1! = 2种,最后一位1为最小的数,故比其小的排列为0。综上,可参考我们常用的十进制和二进制的转换,对于[2, 4, 1], 可总结出其排列的index2! * (2 - 1) + 1! * (3 - 1) + 0! * (1 - 1) + 1.
+

以上分析看似正确无误,实则有个关键的漏洞,在排定第一个数2后,第二位数只可为1或者4,而无法为2,故在计算最终的 index 时需要动态计算某个数的相对大小。按照从低位到高位进行计算,我们可通过两重循环得出到某个索引处值的相对大小。
注意 index 和 factor 的初始值,rank 的值每次计算时都需要重新置零,index 先自增,factorial 后自乘求阶乘。
    public long permutationIndex(int[] A) {
        if (A == null || A.length == 0) return 0L;

        long index = 1, fact = 1;
        for (int i = A.length - 1; i >= 0; i--) {
            // get rank in every iteration
            int rank = 0;
            for (int j = i + 1; j < A.length; j++) {
                if (A[i] > A[j]) rank++;
            }

            index += rank * fact;
            fact *= (A.length - i);
        }

        return index;
    }
http://www.cnblogs.com/EdwardLiu/p/5104310.html
在计算最终的 index 时需要动态计算某个数的相对大小。我们可通过两重循环得出到某个索引处值的相对大小。

以4,1,2为例,4为第3大数,1为剩余序列第1大数,2为剩余序列第1大数,

故表达式为:(3-1)*2! + (1-1)*1! + (1-1)*0! + 1 = 5

以2,4,1为例,2为第2大数,4为剩余序列第2大数,1为剩余序列第1大数

故表达式为:(2-1)*2! + (2-1)*1! + (1-1)*0! + 1 = 4

这后面这个1一定要加,因为前面算的都是比该数小的数,加上这个1,才是该数是第几大数。

2!表示当时当前位后面还有两位,全排列有2!种
 6     public long permutationIndex(int[] A) {
 7         // Write your code here
 8         long res = 0;
 9         int n = A.length;
10         long fact = 1;
11         for (int i=1; i<n; i++) {
12             fact *= i; //fact should at last equal (n-1)!
13         }
14         int initial = n-1; //use to update factorial
15         for (int i=0; i<n; i++) {
16             long count = 0;
17             for (int j=i; j<n; j++) {
18                 if (A[i] >= A[j]) {
19                     count++;
20                 }
21             }
22             res += (count-1)*fact;
23             if (initial != 0) {
24                 fact /= initial;
25                 initial--;
26             }
27         }
28         res = res + 1;
29         return res;
30     }

https://segmentfault.com/a/1190000004683277
搞一个哈希表,存储数组A中每一位A[i]的后面小于它的数的个数count。
为什么这样做呢,因为按照lexicographical order,小的数应该排在前面。那么A[i]后面小于A[i]的数有count个,而i前面又应该有n-i-1位,有(n-1-i)的阶乘种排列的可能,所以应该排在A[i]之前的可能排列就有count * (n-1-i)!个:所以遍历A[]中每一个数,计算在其之前的自然排列的数目,这些数目相加之和存入res,那么res的下一个数字就是现在数组A[]的排列。
对题目有了思索和理解之后,可以找找简化一点的办法。有没有可能不使用HashMap,也能够记录阶乘呢?只要从最后一位fact = 1开始, 向高位阶乘,直到最高位fact = A.length!
    public long permutationIndex(int[] A) {
        int n = A.length;
        long res = 0;
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = i+1; j < n; j++) {
                if (A[i] > A[j]) count++;
            }
            map.put(A[i], count);
        }
        for (int i = 0; i < n; i++) {
            long fact = 1;
            for (int j = n-1-i; j > 0; j--) {
                fact *= j;
            }
            res += map.get(A[i]) * fact;
        }
        return ++res;
    }
    public long permutationIndex(int[] A) {
        if (A == null || A.length == 0) return 0;
        long fact = 1, index = 0;
        for (int i = A.length-1; i >= 0; i--) {
            int rank = 0;
            for (int j = i+1; j < A.length; j++) {
                if (A[j] < A[i]) rank++;
            }
            index += rank * fact;
            fact *= (A.length-i);
        }
        return index+1;
    }
//这种思路是从高位向低位计算当前位剩余排列总数,阶乘逐位减小,理解起来就没有那么直观了。
    public long permutationIndex(int[] A) {
        if (A == null || A.length == 0) return 0;
        long index = 0, fact = 1;
        for (int i = 1; i <= A.length; i++) fact *= i;
        for (int i = 0; i < A.length; i++) {
            int rank = 0;
            for (int j = i+1; j < A.length; j++) {
                if (A[j] < A[i]) rank++;
            }
            fact /= (A.length-i);
            index += rank * fact;
        }
        return index+1;
    }
http://www.jiuzhang.com/solutions/permutation-index/
 long fac(int numerator) {
   
  long now = 1;
  for (int i = 1; i <= numerator; i++) {
   now *= (long) i;
  }
  return now;
 }

 long generateNum(HashMap<Integer, Integer> hash) {
  long denominator = 1;
  int sum = 0;
  for (int val : hash.values()) {
   if(val == 0 ) 
    continue;
   denominator *= fac(val);
   sum += val;
  }
  if(sum==0) {
   return sum;
  }
  return fac(sum) / denominator;
 }

 public long permutationIndex(int[] A) {
  HashMap<Integer, Integer> hash = new HashMap<Integer, Integer>();
  
  for (int i = 0; i < A.length; i++) {
   if (hash.containsKey(A[i]))
    hash.put(A[i], hash.get(A[i]) + 1);
   else {
    hash.put(A[i], 1);
   }
  }
  long ans = 0;
  for (int i = 0; i < A.length; i++) {
   for (int j = i + 1; j < A.length; j++) {
    if (A[j] < A[i]) {
     hash.put(A[j], hash.get(A[j])-1);
     ans += generateNum(hash);
     hash.put(A[j], hash.get(A[j])+1);
     
    }
   
   }
    hash.put(A[i], hash.get(A[i])-1);
  }
  
  return ans+1;

 }
http://www.tuwenzhai.com/d/94558250-815c-4b50-95f3-68d2ed883cd1/ab63e816-26ce-4339-bf94-dd4c325bb3ea


X. DFS
http://lintcode.peqiu.com/content/lintcode/permutation_index.html



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