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
基本思想:數組
這兒由於數組是反向遍歷的,因此可以使用使用stl中的反向迭代器。markdown
runtime:12msui
class Solution {
public:
void nextPermutation(vector<int>& nums) {
if(nums.size()<2) return ;
auto iter=nums.rbegin();
while(iter!=(nums.rend()-1)&&*iter<=*(iter+1))
iter++;
if(iter==nums.rend()-1)
sort(nums.begin(),nums.end());
else
{
auto upper=iter;
auto tmp=nums.rbegin();
for(;tmp!=iter;tmp++)
{
if(*tmp>*(iter+1))
{
if(*tmp<*upper)
{
upper=tmp;
}
}
}
swap(*(iter+1),*upper);
sort(nums.rbegin(),iter+1,greater<int>());
}
}
};