解法1:利用数组记录已访问过的元素
import java.util.*; public class Solution { ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>(); public ArrayList<ArrayList<Integer>> permute(int[] nums) { if (nums == null || nums.length < 1) return res; Arrays.sort(nums); ArrayList<Integer> list = new ArrayList<Integer>(); solve(list, nums); return res; } private void solve(ArrayList<Integer> list, int[] nums) { if (list.size() == nums.length) { res.add(new ArrayList<Integer>(list)); return; } for (int i = 0; i < nums.length; i++) { if (!list.contains(nums[i])) { //如果已经包含就不再加入---相当于剪纸啊? list.add(nums[i]); solve(list, nums); list.remove(list.size() - 1); } } } }
解法2:利用交换
import java.util.*; public class Solution { ArrayList<ArrayList<Integer>> res=new ArrayList<ArrayList<Integer>>(); public ArrayList<ArrayList<Integer>> permute(int[] nums) { if (nums == null || nums.length < 1) return res; Arrays.sort(nums); process1(nums,0); return res; } public void process1(int[] chs,int i){ if(i==chs.length){ ArrayList<Integer>list=new ArrayList<Integer>(); for(int a : chs){ list.add(a); } res.add(list); return; } for(int j=i;j<chs.length;j++){ swap(chs,i,j); process1(chs,i+1); } } public static void swap(int[] chs, int i, int j) { int tmp = chs[i]; chs[i] = chs[j]; chs[j] = tmp; } }