從後往前按照字典升序數，碰到不是升序的某個數字（i）則和從後往前數大於i的最小值（j）交換，然後i+1到陣列末尾的數逆序.如果從尾到頭都是字典升序，則整個陣列逆序

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  and use only constant 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

code：

 public void nextPermutation(int[] nums) {
        if(nums.length<2)
            return;
        
        int i;
        int j=-1;
        StringBuilder builder=new StringBuilder();
        for(i=nums.length-2;i>=0;i--){
            if(nums[i]<nums[i+1]){
                for(j=nums.length-1;j>i;j--){
                    if(nums[j]>nums[i]){
                       swap(nums,i,j);
                        break;
                    }
                }
                int left=i+1;
                int right=nums.length-1;
                while(left<right){
                    swap(nums,left++,right--);
                }
                break;
            }
        }
        if(j==-1){
            int left=0,right=nums.length-1;
            while(left<right)
                swap(nums,left++,right--);
        }
        
    }
    void swap(int[] nums,int i,int j){
        int temp=nums[i];
        nums[i]=nums[j];
        nums[j]=temp;
    }