Next Permutation II(下一个排列)

问题

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).

Have you met this question in a real interview? Yes
Example
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

分析

请参阅 Next Permutation(下一个排列)。

代码

public class Solution {
    /*
     * @param nums: An array of integers
     * @return: nothing
     */
    public void nextPermutation(int[] nums) {
        // write your code here
        for (int i = nums.length - 2; i >= 0; i--) {
            if (nums[i + 1] > nums[i]) {
                for (int j = nums.length - 1; j > i; j--) {
                    if (nums[j] > nums[i]) {
                        int temp = nums[i];
                        nums[i] = nums[j];
                        nums[j] = temp;
                        reverse(nums, i + 1, nums.length-1);
                        return ;
                    }
                }
            }
        }
        reverse(nums, 0, nums.length - 1);
        return ;
    }
    private void reverse(int[] nums, int start, int end) {
        for (int i = start, j = end; i < j; i++, j--) {
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }
}