Permutation Sequence

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Given n and k, return the k-th permutation sequence.

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Example
For n = 3, all permutations are listed as follows:

"123"
"132"
"213"
"231"
"312"
"321"
If k = 4, the fourth permutation is "231"

Note

n will be between 1 and 9 inclusive.

Challenge

O(n*k) in time complexity is easy, can you do it in O(n^2) or less?

Tags Expand

Permutation Array

思路

同样先通过举例来获得更好的理解。以n = 4，k = 9为例：

1234
1243
1324
1342
1423
1432
2134
2143
2314  <= k = 9
2341
2413
2431
3124
3142
3214
3241
3412
3421
4123
4132
4213
4231
4312
4321

最高位可以取{1, 2, 3, 4}，而每个数重复3! = 6次。所以第k=9个permutation的s[0]为{1, 2, 3, 4}中的第9/6+1 = 2个数字s[0] = 2。

而对于以2开头的6个数字而言，k = 9是其中的第k' = 9%(3!) = 3个。而剩下的数字{1, 3, 4}的重复周期为2! = 2次。所以s[1]为{1, 3, 4}中的第k'/(2!)+1 = 2个，即s[1] = 3。

对于以23开头的2个数字而言，k = 9是其中的第k'' = k'%(2!) = 1个。剩下的数字{1, 4}的重复周期为1! = 1次。所以s[2] = 1.

对于以231开头的一个数字而言，k = 9是其中的第k''' = k''/(1!)+1 = 1个。s[3] = 4

public class Solution {
    public String getPermutation(int n, int k) {

        // initialize all numbers
        ArrayList<Integer> numberList = new ArrayList<Integer>();
        for (int i = 1; i <= n; i++) {
            numberList.add(i);
        }

        // change k to be index
        k--;

        // set factorial of n
        int mod = 1;
        for (int i = 1; i <= n; i++) {
            mod = mod * i;
        }

        String result = "";

        // find sequence
        for (int i = 0; i < n; i++) {
            mod = mod / (n - i);
            // find the right number(curIndex) of
            int curIndex = k / mod;
            // update k
            k = k % mod;

            // get number according to curIndex
            result += numberList.get(curIndex);
            // remove from list
            numberList.remove(curIndex);
        }

        return result.toString();
    }
}