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

思路：下一個排列，主要依靠字典序來排列的。首先，從最尾端開始往前尋找兩個相鄰元素，知足num[i]<num[i+1],記錄i，找到這樣的一組相鄰元素，再從最尾端往前查找，找出第一個大於num[i]的元素，索引記爲j。將num[i]和num[j]對調，而後將i+1後的全部元素調到排列。記爲「下一個」排列組合。

class Solution {
public:
    void nextPermutation(vector<int> &num) {
        int n=num.size();
        if(n<=1)
            return;
        int i=n-2;
        while(num[i]>=num[i+1]&&i>=0)
            i--;
        if(i>=0)
        {
            int j=n-1;
            while(num[i]>=num[j]&&j>=0)
                j--;
            swap(num[i],num[j]);
            reverse(num.begin()+i+1,num.end());
        }
        else
        {
            reverse(num.begin(),num.end());
        }
    }
};