Difficulty: Medium
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:
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| Input: n = 3, k = 3
Output: "213"
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Example 2:
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| Input: n = 4, k = 9
Output: "2314"
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Solution
Language: Java
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| class Solution {
public String getPermutation(int n, int k) {
if (n == 0) {
return "";
}
if (n == 1) {
return "1";
}
int fact = 1;
int end = n - 1;
for (int i = 2; i <= end; i++) {
fact *= i;
}
StringBuilder sb = new StringBuilder();
List<Integer> list = new ArrayList<>(n);
for (int i = 1; i <= n; i++) {
list.add(i);
}
k--;
while (sb.length() != n) {
int index = k / fact;
sb.append(list.get(index));
list.remove(index);
k %= fact;
if (end != 0) {
fact /= end;
}
end--;
}
return sb.toString();
}
}
|