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 → 1,3,23,2,1 → 1,2,31,1,5 → 1,5,1
思路：
首先，从后向前找到第一对递增的的数 如 1 4 2 5 7 6 3 中的 5 7， 这表示，7 6 3 是一个连续递减的序列，即这三个数字可以组成的最大的数。 把这三个数翻转，就是 3 6 7即这三个数字可以组成的最小的数字。把5与这三个数字中比它大的最小的数字与它交换变成 1 4 2 6 7 5 3，即下一个较大的数字。注意，像2 3 1 3 3 这样后面比交换点数大一点的数有相同的时候，后面翻转后，交换后面的那个。
class Solution {
public:
void nextPermutation(vector<int> &num)
{
if (num.empty()) return;
// in reverse order, find the first number which is in increasing trend (we call it violated number here)
int i;
for (i = num.size()-; i >= ; --i)
{
if (num[i] < num[i+]) break;
}
// reverse all the numbers after violated number
reverse(begin(num)+i+, end(num));
// if violated number not found, because we have reversed the whole array, then we are done!
if (i == -) return;
// else binary search find the first number larger than the violated number
auto itr = upper_bound(begin(num)+i+, end(num), num[i]);
// swap them, done!
swap(num[i], *itr);
}
};
我自己写得：思路是先交换，再翻转。比上面的繁琐一点。
void nextPermutation(vector<int> &num) {
if(num.empty()) return;
bool b = false;
int chg1 = , chg2 = ;
int post = num.size() - ;
for(int i = num.size() - ; i >= ; --i)
{
if(num[post] > num[i])
{
b = true;
chg1 = i;
chg2 = post;
break;
}
post = i;
}
if(!b)
{
reverse(num.begin(), num.end());
return;
}
for(int i = chg1 + ; i < num.size(); i++)
{
if(num[i] > num[chg1] && num[i] <= num[chg2])
chg2 = i;
}
swap(num[chg1], num[chg2]);
int l1 = chg1 + ;
int l2 = num.size() - ;
while(l1 < l2)
{
swap(num[l1++], num[l2--]);
}
return;
}