[LintCode] Permutation Sequence

Problem

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

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

做法：先把这n个数放入一个数组nums里，同时计算出n的阶乘fact。
然后我们去建立第k个数，也就是java计数规则里的第k-1个数，所以先k--。
怎么建立第k个数呢？这个数有n位数字，所以用0到n-1的for循环来做。
这里应用了一个规律，确定第一个数，有n种选择，每种选择有(n-1)!种情况。选定第一个数之后，选择第二个数，有n-1种选择，每种选择有(n-2)!种情况。选定了前两个数，选择第三个数，有n-2种选择，每种选择有(n-3)!种情况。这样，总共有n!个数，每层循环的样本减少为fact/(n-i)。
所以我们找第k个数，可以先确定它的第一位，从前往后类推。
怎么确定第1位？如上所说，有n种选择，也就是将所有情况分为n组，每种包含(n-1)!个成员。那么，第k个数除以(n-1)!就可以得到这个数在第几组。假设这一组是第m组，第一个数就是nums.get(m)，同时删去这个数，并让k除以(n-1)!取余作为新的k。
之后，把这个数从nums里删去，这样剩余n-1个数的相对位置不变，然后在这一组里找新的第k个数。如此循环，这样，下一组的成员数减少了，要找的位置k也更为精确了。

Some Examples

//1234, n = 4, k = 15,

k = 14, fact = 24,
fact = 24/4 = 6, cur = k/fact = 14/6 = 2, k = k%fact = 2,
nums.get(cur) = 3,
fact = 6/3 = 2, cur = k/fact = 2/2 = 1, k = k%fact = 0,
nums.get(cur) = 2,
fact = 2/2 = 1, cur = k/fact = 0, k = k%fact = 0,
nums.get(0) = 1,
therefore, 3214.

//1234, n = 4, k = 7,

k = 6, fact = 24,
fact = 24/4 = 6, cur = k/fact = 1, k = k%fact = 0,
nums.get(cur) = 2,
fact = 6/3 = 2, cur = 0, nums.get(0) = 1,
cur = 0, nums.get(0) = 3,
cur = 0, nums.get(0) = 4,
therefore, 2134.

//1234, n = 4, k = 22,

k = 21, fact = 24,
fact = 24/4 = 6, cur = k/fact = 3, k = k%fact = 3, 
nums.get(3) = 4,
fact = 2, cur = 3/2 = 1, k = 3%2 = 1, 
nums.get(1) = 2,
fact = 1, cur = 1/1 = 1, k = 1%1 = 0,
nums.get(0) = 3,
fact = 1, cur = 0/1 = 0, k = 0%1 = 0,
nums.get(0) = 1,
therefore, 4231.

Solution

class Solution {
    public String getPermutation(int n, int k) {
        ArrayList<Integer> nums = new ArrayList<Integer>();
        int fact = 1;
        for (int i = 1; i <= n; i++) {
            nums.add(i);
            fact *= i;
        }
        k--;
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < n; i++) {
            fact /= (n-i);
            int cur = k / fact;
            k %= fact;
            sb.append(nums.get(cur));
            nums.remove(cur);
        }
        return sb.toString();
    }
}