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

从后往前，找到第一个i的值比i+1值小的位置。

寻找i以后比i大的最小的数，和i交换。

将i以后的数从小到大排序。

    public void nextPermutation(int[] num) {
        if (num != null && num.length != 0 && num.length != 1) {
            int len = num.length;
            int i = len;
            for (; i > 1; i--) {
                if (num[i - 1] > num[i - 2]) {
                	
                	int flag = 0;
                	
                	for (int k = i - 1; k < len; k++) {
                		if (num[k] <= num[i - 2]) {
                			flag = k - 1;
                			break;
                		}
                	}
                	if (flag == 0) {
                		flag = len - 1;
                	}
                	int temp = num[flag];
                	num[flag] = num[i - 2];
                	num[i - 2] = temp;
                	Arrays.sort(num, i - 1, len);
                	break;
                }              
            }
            if (i == 1) {
                for (int k = 0; k < len / 2; k++) {
                    int temp = num[k];
                    num[k] = num[len - 1 - k];
                    num[len - 1 - k] = temp;
                }
            }
        }
    }