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Permutations.java
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74 lines (69 loc) · 1.96 KB
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/*
46. Permutations
Medium
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 nums are unique.
*/
//CODE 1:
class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
List<Integer> subres = new ArrayList<>();
Solve(nums,subres,res);
return res;
}
public static void Solve(int nums[],List<Integer> subres,List<List<Integer>> res){
if(subres.size()==nums.length){
res.add(new ArrayList<Integer>(subres));
return;
}else{
for(int i=0;i<nums.length;i++){
if(subres.contains(nums[i])) continue;
subres.add(nums[i]);
Solve(nums,subres,res);
subres.remove(subres.size()-1);
}
}
}
}
// CODE 2:
class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
Solve(nums,0,res);
return res;
}
public static void Solve(int nums[],int idx,List<List<Integer>> res){
if(idx==nums.length){
List<Integer> nn = new ArrayList<>();
for(int i=0;i<nums.length;i++){
nn.add(nums[i]);
}
res.add(new ArrayList<Integer>(nn));
return;
}else{
for(int i=idx;i<nums.length;i++){
swap(i,idx,nums);
Solve(nums,idx+1,res);
swap(i,idx,nums);
}
}
}
public static void swap(int i,int j,int[] nums){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}