LeetCode | Next Permutation

題目

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

從後向前（升序）找到第一個逆序對，索引值較小的那個即爲第一個要交換的元素；

再從後向前找到第一個比上述待交換元素大的元素，即爲第二個要交換的元素；

交換兩元素，並將升序序列反轉。

代碼

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;
	}
}