60. Permutation Sequence
摘要：
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 follo...
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"
難度：medium
題目：集合[1, 2, 3, .... n]包含n!個不同的排列。有序的列出所有排列
給定n和k, 返回第k個序列。
思路：結合next_permutation。此方法非簡單方法。
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;
}
}