Table of Contents
Permutations – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
Example 1:
Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Example 2:
Input: nums = [0,1] Output: [[0,1],[1,0]]
Example 3:
Input: nums = [1] Output: [[1]]
Constraints:
1 <= nums.length <= 6-10 <= nums[i] <= 10- All the integers of
numsare unique.
C++ Permutations LeetCode Solution
class Solution {
public:
vector<vector<int> > permute(vector<int> &num) {
vector<vector<int> > result;
permuteRecursive(num, 0, result);
return result;
}
// permute num[begin..end]
// invariant: num[0..begin-1] have been fixed/permuted
void permuteRecursive(vector<int> &num, int begin, vector<vector<int> > &result) {
if (begin >= num.size()) {
// one permutation instance
result.push_back(num);
return;
}
for (int i = begin; i < num.size(); i++) {
swap(num[begin], num[i]);
permuteRecursive(num, begin + 1, result);
// reset
swap(num[begin], num[i]);
}
}
};
Java Permutations LeetCode Solution
public List<List<Integer>> permute(int[] nums) {
if (nums == null || nums.length == 0)
return new ArrayList<>();
List<List<Integer>> finalResult = new ArrayList<>();
permuteRecur(nums, finalResult, new ArrayList<>(), new boolean[nums.length]);
return finalResult;
}
private void permuteRecur(int[] nums, List<List<Integer>> finalResult, List<Integer> currResult, boolean[] used) {
if (currResult.size() == nums.length) {
finalResult.add(new ArrayList<>(currResult));
return;
}
for (int i = 0; i < nums.length; i++) {
if (used[i])
continue;
currResult.add(nums[i]);
used[i] = true;
permuteRecur(nums, finalResult, currResult, used);
used[i] = false;
currResult.remove(currResult.size() - 1);
}
}
Python 3 Permutations LeetCode Solution
# DFS
def permute(self, nums):
res = []
self.dfs(nums, [], res)
return res
def dfs(self, nums, path, res):
if not nums:
res.append(path)
# return # backtracking
for i in xrange(len(nums)):
self.dfs(nums[:i]+nums[i+1:], path+[nums[i]], res)
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
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