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

题目：实现 下一个排列，其按字典序重新排列给定数组为比下一个排列更大的排列。

思路：下一个排列实际上比当前排列大且最接近的排列。具体做法是：在数组中从后往前，找到最后升序的地方，交换前后值，并将后面的原序逆转为升序。

	public void nextPermutation(int[] num) {
		int len = num.length, i = 0, j = 0;
		for (i = len - 2; i >= 0; i--) {
			if (num[i] >= num[i + 1])
				continue;
			for (j = len - 1; j > i; j--) {
				if (num[j] > num[i])
					break;
			}
			break;
		}
		if (i >= 0) {
			int temp = num[i];
			num[i] = num[j];
			num[j] = temp;
		}
		int end = len - 1;
		int start = i + 1;
		while (start < end) {
			int temp = num[start];
			num[start] = num[end];
			num[end] = temp;
			start++;
			end--;
		}
	}