Given a collection of distinct integers, return all possible permutations.
Example:
Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
public class Permutations { public List<List<Integer>> permute(List<Integer> nums) { List<List<Integer>> result = new ArrayList<>(); result.add(new ArrayList<>()); for(int num: nums) { int sz = result.size(); for(int i = 0; i < sz; ++i) { List<Integer> old = result.get(i); for(int j = 0; j <= old.size(); ++j) { List<Integer> updated = new ArrayList<>(old); updated.add(j, num); result.add(updated); } } result.subList(0, sz).clear(); } return result; } public static void main(String[] args) { Permutations p = new Permutations(); List<List<Integer>> result = p.permute(Arrays.asList(1, 2, 3)); System.out.println(result); } }
Time complexity: O(n * n!) = O(0! + 1! + 2! + 3! + ...n!)
Iterative: staring with an empty array, { { } }
Adding number 1, { { } } ->{ { 1 } }
Adding number 2, { { 1 } } ->{ { 2 , 1 }, { 1, 2 } }
Adding number 3: { { 2, 1 }, { 1, 2 } } -> { {3, 2, 1}, {2, 3, 1}, {2, 1, 3}, {3, 1, 2}, {1, 3, 2}, {1, 2, 3}}
Taken-out: start with something small, build the solution based on smaller inputs.
Recursive: backtracking, swapping elements to get new array
1. {1, 2, 3} swap the first element with the rest, i = 0
2. After 1 swapped with itsself, {1, 2, 3}, swap the 2nd element with the rest, pos = 1
3. After 2 swapped with itselft, {1, 2, 3}, swap the 3rd element with the rest, pos = 2
4. After 3 swapped with iteself, {1, 2, 3}, you can either return here when pos == nums.size() or pos > nums.size(), result.add(new ArrayList<>(nums)), as base case. {{1, 2, 3}}
5. Back to step 3, 2 swapped with 3, {1, 3, 2}, swap the 3rd element with the rest, pos = 2
6. After 2 swapped with iteself, {1, 3, 2}. { {1, 2, 3}, {1, 3, 2} }
7. Backtracked to step 3, after the subcall which swapped 2 and 3, in order to return the nums to previous state before the subcall, we need to swap the elemtns back.
public class Permutations { private void permuteHelper(int pos, List<Integer> nums, List<List<Integer>> result) { int len = nums.size(); if(pos == len-1) { result.add(new ArrayList<>(nums)); return; } for(int i = pos; i < len; ++i) { Collections.swap(nums, pos, i); permuteHelper(pos+1, nums, result); Collections.swap(nums, pos, i); } } public List<List<Integer>> permute(List<Integer> nums) { List<List<Integer>> result = new ArrayList<>(); permuteHelper(0, nums, result); return result; } public static void main(String[] args) { Permutations p = new Permutations(); List<List<Integer>> result = p.permute(Arrays.asList(1, 2, 3)); System.out.println(result); } }
原文：https://www.cnblogs.com/taste-it-own-it-love-it/p/10352376.html