Next Permutation – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
A permutation of an array of integers is an arrangement of its members into a sequence or linear order.
- For example, for
arr = [1,2,3]
, the following are all the permutations ofarr
:[1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1]
.
The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).
- For example, the next permutation of
arr = [1,2,3]
is[1,3,2]
. - Similarly, the next permutation of
arr = [2,3,1]
is[3,1,2]
. - While the next permutation of
arr = [3,2,1]
is[1,2,3]
because[3,2,1]
does not have a lexicographical larger rearrangement.
Given an array of integers nums
, find the next permutation of nums
.
The replacement must be in place and use only constant extra memory.
Example 1:
Input: nums = [1,2,3] Output: [1,3,2]
Example 2:
Input: nums = [3,2,1] Output: [1,2,3]
Example 3:
Input: nums = [1,1,5] Output: [1,5,1]
Constraints:
1 <= nums.length <= 100
0 <= nums[i] <= 100
C++ Next Permutation LeetCode Solution
void nextPermutation(vector<int>& nums) {
int i = nums.size() - 1, k = i;
while (i > 0 && nums[i-1] >= nums[i])
i--;
for (int j=i; j<k; j++, k--)
swap(nums[j], nums[k]);
if (i > 0) {
k = i--;
while (nums[k] <= nums[i])
k++;
swap(nums[i], nums[k]);
}
}
Java Next Permutation LeetCode Solution
public void nextPermutation(int[] A) {
if(A == null || A.length <= 1) return;
int i = A.length - 2;
while(i >= 0 && A[i] >= A[i + 1]) i--; // Find 1st id i that breaks descending order
if(i >= 0) { // If not entirely descending
int j = A.length - 1; // Start from the end
while(A[j] <= A[i]) j--; // Find rightmost first larger id j
swap(A, i, j); // Switch i and j
}
reverse(A, i + 1, A.length - 1); // Reverse the descending sequence
}
public void swap(int[] A, int i, int j) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
public void reverse(int[] A, int i, int j) {
while(i < j) swap(A, i++, j--);
}
Python 3 Next Permutation LeetCode Solution
def nextPermutation(self, nums):
i = j = len(nums)-1
while i > 0 and nums[i-1] >= nums[i]:
i -= 1
if i == 0: # nums are in descending order
nums.reverse()
return
k = i - 1 # find the last "ascending" position
while nums[j] <= nums[k]:
j -= 1
nums[k], nums[j] = nums[j], nums[k]
l, r = k+1, len(nums)-1 # reverse the second part
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l +=1 ; r -= 1
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
Leave a comment below