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 and use only constant 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
解法一
class Solution {
public:
void nextPermutation(vector<int>& nums) {
int i = nums.size()-2;
while(i>=0&&nums[i+1]<=nums[i]){
i--;
}
if(i>=0){
int j=nums.size()-1;
while(j>=0&&nums[j]<=nums[i])
j--;
swap(nums[i],nums[j]);
}
reverse(nums,i+1);
}
void reverse(vector<int>& nums,int start){
int i=start,j=nums.size()-1;
while(i<j){
swap(nums[i],nums[j]);
i++;
j--;
}
}
};
解法二
class Solution {
public:
void nextPermutation(vector<int> &nums) {
next_permutation(nums.begin(), nums.end());
}
template<typename BidiIt> bool next_permutation(BidiIt first, BidiIt last) {
// Get a reversed range to simplify reversed traversal.
const auto rfirst = reverse_iterator<BidiIt>(last);
const auto rlast = reverse_iterator<BidiIt>(first);
// Begin from the second last element to the first element.
auto pivot = next(rfirst);
// Find `pivot`, which is the first element that is no less than its // successor. `Prev` is used since `pivort` is a `reversed_iterator`.
while (pivot != rlast && *pivot >= *prev(pivot))
++pivot;
// No such elemenet found, current sequence is already the largest
// permutation, then rearrange to the first permutation and return false.
if (pivot == rlast) {
reverse(rfirst, rlast);
return false;
}
// Scan from right to left, find the first element that is greater than // `pivot`.
auto change = find_if(rfirst, pivot, bind1st(less<int>(), *pivot));
swap(*change, *pivot);
reverse(rfirst, pivot);
return true;
}
};
