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 for n = 3:
"123" "132" "213" "231" "312" "321"
Given n and k, return the kth permutation sequence.
Note:
Given n will be between 1 and 9 inclusive. Given k will be between 1 and n! inclusive.
Example 1:
Input: n = 3, k = 3 Output: "213"
Example 2:
Input: n = 4, k = 9 Output: "2314"
难度:medium
题目:集合[1, 2, 3, .... n]包含n!个不同的排列。有序的列出所有排列
给定n和k, 返回第k个序列。
思路:结合next_permutation。此方法非简单方法。
Runtime: 31 ms, faster than 16.21% of Java online submissions for Permutation Sequence.
Memory Usage: 37.6 MB, less than 6.50% of Java online submissions for Permutation Sequence.
class Solution { public String getPermutation(int n, int k) { int[] nums = new int[n]; for (int i = 0; i < n; i++) { nums[i] = i + 1; } for (int i = 0; i < k - 1; i++) { nextPermutation(nums); } StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { sb.append(nums[i]); } return sb.toString(); } public void nextPermutation(int[] nums) { if (null == nums || nums.length <= 1) { return; } int right = nums.length - 1, j = right; for (; j > 0; j--){ if (nums[j - 1] < nums[j]) break; } for (int k = right; k >= 0; k--) { if (j > 0 && nums[k] > nums[j - 1]) { swap(nums, j - 1, k); break; } } for (; j < right; j++, right--) { swap(nums, j, right); } } private void swap(int[] nums, int i, int j) { int t = nums[i]; nums[i] = nums[j]; nums[j] = t; } }