LeetCode | Next Permutation

题目

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

分析

从后向前（升序）找到第一个逆序对，索引值较小的那个即为第一个要交换的元素；

再从后向前找到第一个比上述待交换元素大的元素，即为第二个要交换的元素；

交换两元素，并将升序序列反转。

代码

public class NextPermutation {
	public void nextPermutation(int[] num) {
		int i = num.length - 2;
		// find the cut point
		while (i >= 0 && num[i] >= num[i + 1]) {
			--i;
		}

		// find the swap point
		if (i >= 0) {
			int j = num.length - 1;
			while (num[j] <= num[i]) {
				--j;
			}
			swap(num, i, j);
		}

		// reverse
		reverse(num, i + 1);
	}

	private void reverse(int[] num, int start) {
		int end = num.length - 1;
		while (start < end) {
			swap(num, start, end);
			++start;
			--end;
		}
	}

	private void swap(int[] num, int i, int j) {
		int tmp = num[i];
		num[i] = num[j];
		num[j] = tmp;
	}
}