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

秒速五厘米 2022-05-23 11:39 14阅读 0赞

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;
        }

[in-place]: http://en.wikipedia.org/wiki/In-place_algorithm