Friday, April 15, 2016

Lintcode 198 - Permutation Index II


http://www.cnblogs.com/EdwardLiu/p/5104409.html
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.

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

这里需要考虑重复元素,有无重复元素最大的区别在于原来的1!, 2!, 3!...等需要除以重复元素个数的阶乘。记录重复元素个数同样需要动态更新,引入哈希表这个万能的工具较为方便。

按照从数字低位到高位进行计算
 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     }
http://blog.csdn.net/earthma/article/details/47074779
这道题基于查找不存在重复元素中排列序号的基础之上,
即P(n) = P(n-1)+C(n-1)
C(n-1) = (首元素为小于当前元素,之后的全排列值)
P(1) = 1;
而不存在重复元素的全排列值C(n-1) = (n-1)!*k(k为首元素之后小于当前元素的个数)
在存在重复元素的排列中首先全排列的值的求法变为:
C(n-1) = (n-1)!/(A1!*A2!*···*Aj!)*k(其中Ai 为重复元素的个数,k为小于首元素前不重复的个数)

http://algorithm.yuanbin.me/zh-hans/exhaustive_search/permutation_index_ii.html
这一题有了重复元素,就是要去掉重复的排列,其实很想排列组合问题,对于重复的数字我们应该去掉他们重复的组合,就是不在乎重复数字的重复排列,应该从原来的结果中除以重复元素个数的阶乘hashmap记录重复元素的个数
注意点
  • 在loop中每遇到一个元素,都要重新建立一个hashmap来计算
  • 元素自己也要先加入到hashmap中,不然在右边计算重复元素的时候就可能少掉自己的一次

这里需要考虑重复元素,有无重复元素最大的区别在于原来的1!, 2!, 3!...等需要除以重复元素个数的阶乘,颇有点高中排列组合题的味道。记录重复元素个数同样需要动态更新,引入哈希表这个万能的工具较为方便。

在计算重复元素个数的阶乘时需要注意更新multiFact的值即可,不必每次都去计算哈希表中的值。对元素A[i]需要加入哈希表 - hash.put(A[i], 1);,设想一下2, 2, 1, 1的计算即可知。
复杂度分析

双重 for 循环,时间复杂度为 O(n^2)O(n​2​​), 使用了哈希表,空间复杂度为 O(n)O(n).
    public long permutationIndexII(int[] A) {
        if (A == null || A.length == 0) return 0L;

        Map<Integer, Integer> hashmap = new HashMap<Integer, Integer>();
        long index = 1, fact = 1, multiFact = 1;
        for (int i = A.length - 1; i >= 0; i--) {
            // collect its repeated times and update multiFact
            if (hashmap.containsKey(A[i])) {
                hashmap.put(A[i], hashmap.get(A[i]) + 1);
                multiFact *= hashmap.get(A[i]);
            } else {
                hashmap.put(A[i], 1);
            }
            // get rank every turns
            int rank = 0;
            for (int j = i + 1; j < A.length; j++) {
                if (A[i] > A[j]) rank++;
            }
            // do divide by multiFact
            index += rank * fact / multiFact;
            fact *= (A.length - i);
        }

        return 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 permutationIndexII(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++) {
   HashMap<Integer, Integer> flag = new HashMap<Integer, Integer>();
   
   for (int j = i + 1; j < A.length; j++) {
    if (A[j] < A[i] && !flag.containsKey(A[j])) {
         flag.put(A[j], 1);
    
     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;

 }




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