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
问题描述：重新排列数字，得到按大小排列的下一个数字，如果没有这样的排列，也就是说不能得到比它还大的，就按照升序重新排列。
要求必须原地进行，不能分配额外空间。
方法是，从后面往前面遍历，假设当前比较的后面的值是a，前面的值是b，如果b>=a，则继续遍历，如果b
=a，则重新按照升序排列。
class Solution {
public:
void nextPermutation(vector &num) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int tmp = 0;
vector::iterator iter, viter;
for(viter = num.end()-1;
viter != num.begin(); --viter) {
if(*viter > *(viter-1)) {
iter = viter;
while(iter != num.end() && *iter > *(viter-1))
++iter;
--iter;
tmp = *iter;
num.erase(iter);
iter = num.insert(viter-1, tmp);
sort(iter+1, num.end());
break;
}
}
if(viter == num.begin()) {
sort(num.begin(), num.end());
}
}
};