LeetCode OJ--Next Permutation *

求一个排列的下一个排列。

1,2,3 → 1,3,2
3,2,1 → 1,2,3
1,1,5 → 1,5,1

#include <iostream>

#include <vector>

#include <algorithm>

using namespace std;

class Solution{

public:

    void nextPermutation(vector<int> &num) {

        if(num.size() == 0)

            return;
const vector<int>::reverse_iterator rfirst= num.rbegin();

        const vector<int>::reverse_iterator rend = num.rend();
auto pivot = next(rfirst);

        while(pivot != rend && *pivot >= *prev(pivot))

        {

            ++pivot;

        }
if(pivot == rend)

        {

            reverse(rfirst,rend);

            return;

        }

        //find the first num great than pivot

        auto change = rfirst;

        while(*change<=*pivot)

            ++change;
swap(*change,*pivot);

        reverse(rfirst,pivot);

        return;

    }

};
int main()

{

    vector<int> num;

    num.push_back(1);

    num.push_back(1);

    num.push_back(5);
Solution myS;

    myS.nextPermutation(num);

    return 0;

}

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