Permutation Index

26% Accepted

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.

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

直接想到的就是把每个都写出来,放进hashmap里边,然后去找hashmap对应的index

然而Memory Limit Exceeded

public class Solution {
    /**
     * @param A an integer array
     * @return a long integer
     */
    public long permutationIndex(int[] A) {
        // Write your code here
        int[] B = new int[A.length];
        for (int i = 0; i< A.length; i++) {
            B[i] = A[i];
        }
        Arrays.sort(B);
        HashMap<ArrayList<Integer>, Integer> hashmap = new HashMap<ArrayList<Integer>, Integer>();
        ArrayList<Integer> permutation = new ArrayList<Integer>();
        dfs(B, hashmap, permutation);
        ArrayList<Integer> Acopy = new ArrayList<Integer>();
        for (int i = 0; i < A.length; i++) {
            Acopy.add(A[i]);
        }
        return hashmap.get(Acopy) + 1;
    }

    private int index = 0;
    public void dfs(int[] B, HashMap<ArrayList<Integer>, Integer> hashmap, ArrayList<Integer> permutation) {
        if (permutation.size() == B.length) {
            hashmap.put(new ArrayList<Integer>(permutation), index++);
            return;
        }
        for (int i = 0; i < B.length; i++) {
            if (permutation.contains(B[i])) {
                continue;
            }
            permutation.add(B[i]);
            dfs(B, hashmap, permutation);
            permutation.remove(permutation.size() - 1);
        }
    }
}

正确解法

思路

1.对于四位数:4213 = 4*100+2*100+1*10+3

2.4个数的排列有4!种。当我们知道第一位数的时候,还有3!种方式,当知道第二位数时候还有2!种方式,当知道第三位数的时候还有1!种方式,前面三位数都确定的时候,最后一位也确定了。<这里是按照高位到地位的顺序>

3.对4个数的排列,各位的权值为:3!,2!,1!,0!。第一位之后的数小于第一位的个数是x,第二位之后的数小于第二位的个数是y,第三位之后的数小于第三的个数是z,第四位之后的数小于第四位的个数是w,则abcd排列所在的序列号:index = x*3!+y*2!+z*1!,<0!=0>

在数的排列中,小数在前面,大数在后面,所以考虑该位数之后的数小于该为的数的个数,这里我自己理解的也不是很透,就这样。

4.例如 4213;x= 3,y = 1,z=0,index = 18+2=20

123;x = 0,y=0,index = 0

321;x= 2,y=1,index = 2*2!+1*1! = 5

这里的下标是从0开始的。
public class Solution {
    /**
     * @param A an integer array
     * @return a long integer
     */
    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;
    }
}

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