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

1.先从右到左找到左面比右面小的第一个数字，记下标为i

2.从右到左面找到比i大，但是最小的数，记小标为j

3，交换i.j数

4.对i右面的数进行反转。

class Solution {
public:
void nextPermutation(vector<int>& nums) {
int n=nums.size();
int i,j;
if(n<=1)
return ;
for(i=n-2;i>=0;--i)
{
if(nums[i]<nums[i+1])
break;
}
if(i=-1)//这个地方出错，应该是i==-1
{
reverse(nums.begin(),nums.end());
return;
}
for(j=n-1;j>i;--j)
{
if(nums[i]<nums[j]){
break;
}

}
swap(nums[i],nums[j]);
reverse(nums.begin()+i+1,nums.end());
}
};