LeetCode 31

原题:（频率2）

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

题意：求下一序列，从小到大，如果没有下一序列，就回到最开始最小的序列。

代码和思路：

class Solution {
    public void nextPermutation(int[] nums) {
        int i = nums.length - 2;
        //从右往左，找到第一个开始递减的数，标记为i
        while(i>=0 && nums[i]>=nums[i+1]){
            i--;
        }
        if(i>=0){
           //再找到一个j，刚好大于i
            int j = nums.length - 1;
            while(j>=0 && nums[j]<=nums[i]){
                j--;
            }
            //交换i，j
            swap(nums,i,j);
        }
        //从i+1到最后倒序
        reverse(nums,i+1);
    }
    
    private void swap(int [] nums,int a,int b){
        int temp = nums[a];
        nums[a] = nums[b];
        nums[b] = temp;
    }
  
    
    private void reverse(int [] nums,int start){
        int end = nums.length - 1;
        while(start < end){
            swap(nums,start,end);
            start++;
            end--;
        }
    }
}