[LeetCode]--60. Permutation Sequence

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本文介绍了一种高效算法来解决求解第K个排列序列的问题，并提供了两种不同的实现方式，一种通过直接生成所有可能的排列，另一种则利用数学原理进行优化。

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1."123"
2."132"
3."213"
4."231"
5."312"
6."321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

[LeetCode]–46. Permutations

前面做过很多个这样类似题，所以很快就能想到找下一个字串。

public String getPermutation(int n, int k) {
        if (n == 1)
            return "1";
        int[] nums = new int[n];
        for (int i = 0; i < n; i++)
            nums[i] = i + 1;
        for (int i = 1; i < k; i++) {
            nums = getNextPermutation(nums);
        }
        String str = "";
        for (int i = 0; i < nums.length; i++) {
            str += nums[i];
        }
        return str;
    }

    public int[] getNextPermutation(int[] nums) {
        int i = nums.length - 1;
        while (i > 0 && nums[i] < nums[i - 1])
            i--;
        int second = Integer.MAX_VALUE, secondIndex = Integer.MAX_VALUE;
        for (int j = nums.length - 1; j >= i; j--)
            if (nums[j] > nums[i - 1] && nums[j] < second) {
                second = nums[j];
                secondIndex = j;
            }
        int temp = nums[i - 1];
        nums[i - 1] = nums[secondIndex];
        nums[secondIndex] = temp;
        Arrays.sort(nums, i, nums.length);
        return nums;
    }

虽然AC了，不过效率实在是很低，我又找到了一个效率很高的，beat88%的算法。

public String getPermutation(int n, int k) {
        StringBuilder sb = new StringBuilder();
        boolean[] used = new boolean[n];

        k = k - 1;
        int factor = 1;
        for (int i = 1; i < n; i++) {
            factor *= i;
        }

        for (int i = 0; i < n; i++) {
            int index = k / factor;
            k = k % factor;
            for (int j = 0; j < n; j++) {
                if (used[j] == false) {
                    if (index == 0) {
                        used[j] = true;
                        sb.append((char) ('0' + j + 1));
                        break;
                    } else {
                        index--;
                    }
                }
            }
            if (i < n - 1) {
                factor = factor / (n - 1 - i);
            }
        }

        return sb.toString();
    }