leetcode刷题系列C++-next permutation

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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class Solution {public:    void nextPermutation(vector<int>& nums) {              int pos = -1;        int length = nums.size();        //找到第一个升序的位置        for (int i = length - 1; i > 0; --i)        {            if(nums[i] > nums[ i - 1])            {                pos = i - 1;                break;            }        }        //如果没有找到 也就是一直都是升序的 pos为负值 那么直接反序即可        if(pos < 0)        {            reverse(nums,0,length - 1);            return;        }        //找到了破坏升序的那个数据 将该元素跟第一个比他大的元素交换        for(int i = length - 1; i > pos; --i)        {            if(nums[i] > nums[pos])            {                int tmp = nums[i];                nums[i] = nums[pos];                nums[pos] = tmp;                break;            }        }        reverse(nums,pos + 1, length - 1);            }    void reverse(vector<int>& nums, int begin,int end)    {        int tmp = 0;        while(begin < end)        {            tmp = nums[begin];            nums[begin] = nums[end];            nums[end] = tmp;            begin++;            end--;        }    }};