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).
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
The replacement must be in-place, do not allocate extra memory.
class Solution {
public:
/**
* @param nums: a vector of integers
* @return: return nothing (void), do not return anything, modify nums in-place instead
*/
void nextPermutation(vector<int> &nums) {
// write your code here
// write your code here
bool finished = false;
if (nums.size() <= 1)
finished = true;
priority_queue<int, vector<int>> pq;
vector<int> back_vec;
pq.push(INT_MIN);
unordered_map<int, int> map;
for (int i = nums.size() - 1; i >= 0 && !finished; i--) {
if (nums[i] < pq.top()) {
back_vec.push_back(nums[i]);
sort(back_vec.begin(), back_vec.end());
auto itr = upper_bound(back_vec.begin(), back_vec.end(), nums[i]);
nums[i] = *itr;
back_vec.erase(itr);
nums.erase(nums.begin() + i + 1, nums.end());
nums.insert(nums.end(), back_vec.begin(), back_vec.end());
finished = true;
} else {
pq.push(nums[i]);
back_vec.push_back(nums[i]);
}
}
if (!finished)
sort(nums.begin(), nums.end());
}
};