Leetcode: Next Permutation

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

思路：从右往左找第一个降序位置，记为a，再从右往左找第一个比a大的位置b，交换这两个位置上的值，并把a之后的数列重新按照大小排列，得到结果。如果从右往左找不到降序，则将整个数列都重新排序。

Solution:

public class Solution {    public void nextPermutation(int[] num) {        if (num == null || num.length == 0 || num.length == 1) {            return;        }                int size = num.length;        for (int i = size - 1; i >= 0; i--) {        if (i == 0) {        Arrays.sort(num);        return;        }        if (num[i] > num[i - 1]) {        int temp1 = num[i - 1];        for (int j = size - 1; j >= i; j--) {        if (temp1 < num[j]) {        num[i - 1] = num[j];        num[j] = temp1;        break;        }         }        Arrays.sort(num, i, size);        return;        }        }    }}