27 - 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

solution:  找出最尾的顺序对, 反转。先从尾往头扫到非递增的数截止（若扫到0，则表示该数最大，立即反转），然后以该数为对比继续从尾往头扫，直到有一个数比他大，交换这两个数，然后立即反转整个从尾递增的数列。

class Solution {public:        void nextPermutation(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(num.size() == 0)            return;        int len = num.size();        int fir = len -1;        for( fir; fir >= 1; fir--)        {            if( num[fir] > num[fir-1] )                break;        }                if( fir > 0 )        {            fir --;            int sec = len -1;            while( sec >= 0 && num[sec] <= num[fir] )            {                sec --;            }            //swap fir and sec            int temp = num[ fir ];            num[ fir ] = num[ sec ];            num[ sec ] = temp;            fir ++;        }                int end = len - 1;        while(end > fir)        {            int temp = num[ fir ];            num[ fir ] = num[ end ];            num[ end ] = temp;            end--;            fir++;        }            }};