lintcode-medium-Permutation Index II

最新推荐文章于 2020-07-12 12:05:10 发布

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最新推荐文章于 2020-07-12 12:05:10 发布

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原文链接：http://www.cnblogs.com/goblinengineer/p/5351817.html

本文介绍了如何在给定重复数字的排列中找到其在字典序排列中的特定索引位置，通过使用哈希表记录每个数字出现的次数，以及递归计算小于当前元素的所有可能排列数量。

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.

public class Solution {
    /**
     * @param A an integer array
     * @return a long integer
     */
    public long permutationIndexII(int[] A) {
        // Write your code here
        
        if(A == null || A.length == 0)
            return 1;
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        
        for(int i = 0; i < A.length; i++){
            if(map.containsKey(A[i]))
                map.put(A[i], map.get(A[i]) + 1);
            else
                map.put(A[i], 1);
        }
        
        long result = 1;
        
        for(int i = 0; i < A.length; i++){
            HashSet<Integer> set = new HashSet<Integer>();
            
            for(int j = i + 1; j < A.length; j++){
                if(A[j] < A[i] && !set.contains(A[j])){
                    set.add(A[j]);
                    
                    map.put(A[j], map.get(A[j]) - 1);
                    result += getNum(map);
                    map.put(A[j], map.get(A[j]) + 1);
                }
            }
            
            map.put(A[i], map.get(A[i]) - 1);
        }
        
        return result;
    }
    
    public long factor(int n){
        long result = 1;
        
        for(int i = 1; i <= n; i++)
            result = result * i;
        
        return result;
    }
    
    public long getNum(HashMap<Integer, Integer> map){
        long den = 1;
        int count = 0;
        
        for(Integer temp: map.values()){
            if(temp != 0){
                count += temp;
                den *= factor(temp);
            }
        }
        
        return factor(count) / den;
    }
    
}