leetcode Next Permutation

Next Permutation

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Total Accepted: 48550 Total Submissions: 193196 Difficulty: Medium

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

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C++的STL库里面有nextPermutation()方法，其实现就是字典序法。

下图简单明了地介绍了字典序法

归纳一下为：

例如，1234的全排列如下：

注意，还有另外一种情况需要特别处理，就是当比该排列大的排列不存在时候，需要反转。

class Solution {public:    void nextPermutation(vector<int>& nums)     {        int right=nums.size()-1;        int len=right;        if(right<1)            return;        int beforeVal=right-1;        int temp;        while(beforeVal>=0)        {            if(nums[right]>nums[beforeVal])            {                break;            }            else            {                right--;                beforeVal--;            }        }        if(beforeVal<0)        {            reserve(nums,0,len);            return;        }        int j=len;        while(j>=0 && nums[j]<=nums[beforeVal])            j--;        swap(nums[j],nums[beforeVal]);        reserve(nums,beforeVal+1,len);         }    void reserve(vector<int>& nums,int i,int j)    {        while(j>i)        {            swap(nums[i],nums[j]);            i++;            j--;        }    }};