Next Permutation

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

The replacement must be in-place, do not allocate extra memory.

Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column.

1,2,3 → 1,3,2

3,2,1 → 1,2,3

1,1,5 → 1,5,1

Solution:

  1. public class Solution {
  2.   public void nextPermutation(int[] nums) {
  3.     if (nums == null || nums.length == 0) {
  4.       return;
  5.     }
  7.     int n = nums.length;
  9.     // step 1. scan from right and find the digit that is lower than the one on its right
  10.     int p = n - 1;
  11.     while (p > 0 && nums[p - 1] >= nums[p]) { p--; }
  13.     if (p == 0) {
  14.       // no such digit is found, the whole array is sorted in descending order
  15.       // we can simply reverse it
  16.       reverse(nums, 0, n - 1);
  17.       return;
  18.     }
  20.     // step 2. from p, find the digit that is just larger than nums[p - 1]
  21.     int i = p - 1;
  22.     int j = p;
  24.     while (p < n) {
  25.       if (nums[p] > nums[i] && nums[p] <= nums[j]) {
  26.         j = p;
  27.       }
  28.       p++;
  29.     }
  31.     // step 3. swap i & j
  32.     swap(nums, i, j);
  34.     // step 4. reverse the digits after i
  35.     reverse(nums, i + 1, n - 1);
  36.   }
  38.   void swap(int[] nums, int i, int j) {
  39.     int t = nums[i];
  40.     nums[i] = nums[j];
  41.     nums[j] = t;
  42.   }
  44.   void reverse(int[] nums, int i, int j) {
  45.     while (i < j) {
  46.       swap(nums, i++, j--);
  47.     }
  48.   }
  49. }