31. 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 

解题思路:

以3457621为例，

1.从尾部寻找第一个满足p(i-1)<p(i)的情况，即

35-7621，5<7 

2.从i-1往回找最小的大于p(i-1)的数minLarge,交换p(i-1),minLarge.  

367521 

3.将i-1后面的数从小到大排序，若是全部是不同的数，只需逆序，

3461257

 void nextPermutation(vector<int>& nums) {        int small=nums[nums.size()-1];        int i,j,minLarge;        bool flag=false;        for(i=nums.size()-1;i>0;i--){            if(nums[i]>nums[i-1])            {                flag=true;                small=nums[i-1];                break;            }        }        if(flag)        {            int big=nums[i];            minLarge=i;            for(j=i;j<nums.size();j++){                if(nums[j]>small)                {                    if(big>nums[j])                    {                        big=nums[j];                        minLarge=j;                    }                    }            }            swap(nums[i-1],nums[minLarge]);            vector<int>::iterator iter=nums.begin()+i;            sort(iter,nums.end());        }        else{            sort(nums.begin(),nums.end());        }    }