leetcode-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：从后往前找，当找到某个元素num[i]>num[i-1]，则把A[i]……A[n]的元素进行排序，

排完序再查找第一个比num[i-1]大的元素，将两个元素进行交换。（主要是发现permutation的特点）

代码：

void nextPermutation(vector<int> &num) 
{
int len=num.size();
int p=0;
if(len>1)
{
int i=len-1;
while(i>0)
{
if(num[i-1] < num[i])
{
if(i == len-1)
{
swap(num[i-1],num[i]);
}
break;
}
else if(num[i-1] == num[i])
{
if(i==1)
{
break;
}
--i;
}
else if(num[i-1] >num[i])
{
--i;
}
}

if(i<(len-1))
{
sort(num.begin()+i,num.end());


p=i;
while(p<len && num[i-1]<num[p])
{
++p;
}
swap(num[i-1],num[p]);
}
}
}