Next Permutation(从后往前字典序列升序，否则交换逆序然后输出。。)leetcode31

本文链接：https://blog.youkuaiyun.com/m0_37402140/article/details/80616860

本文介绍了一个用于生成整数数组下一个字典序排列的算法。该算法首先从数组末尾开始寻找第一个破坏升序排列的元素，并与后续子数组中大于其的最小值交换位置，随后将该元素之后的部分反转以得到下一个排列。若数组完全降序，则将其整体反转。

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从后往前按照字典升序数，碰到不是升序的某个数字（i）则和从后往前数大于i的最小值（j）交换，然后i+1到数组末尾的数逆序.如果从尾到头都是字典升序，则整个数组逆序

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 in-place 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

4,6,7,2,6,5,4,3,0--->4,6,7,3,0,2,4,5,6

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;
    }