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：

1、从末尾到头遍历元素，找到两个相邻的元素a,b,a < b,(a索引为i,b索引为i+1)；

2、第一步若未找到，说明此序列为最大值，则翻转整个序列为最小；

3、若找到了，则从末尾到i+1,查询第一个比a大的元素，与a交换；

4、然后翻转i+1到末尾；

代码：

public class Solution {
    private void swap(int[] num, int i, int j) {
		// TODO Auto-generated method stub
		int tmp = num[i];
		num[i] = num[j]; 
		num[j] = tmp;
	}
	private void reverse(int[] nums, int left, int right){
	     while(left= 0; i--,ii--){
            if(num[i] < num[ii])
            {
                int j = n-1;
                while(num[j] <= num[i])j--;//从尾部找到第一个比num[i]大的数，一定可以找到
                swap(num, i, j);
                reverse(num, ii, n - 1);
                return;
            }
        }
        reverse(num, 0, n - 1);
    }
}