Backtracking（一）

LeetCode中涉及到回溯的题目有通用的解题套路：

46. permutations

这一类回溯题目中的基础中的基础，无重复数字枚举：

 1 /* Given a collection of distinct numbers, return all possible permutations.
 2 
 3 For example,
 4 [1,2,3] have the following permutations:
 5 [
 6   [1,2,3],
 7   [1,3,2],
 8   [2,1,3],
 9   [2,3,1],
10   [3,1,2],
11   [3,2,1]
12 ] */
13 
14 public class Solution {
15     public List<List<Integer>> permute(int[] nums) {
16         List<List<Integer>> res = new ArrayList<>();
17         backtrack(res, new ArrayList<>(), nums);
18         return res;
19     }
20     public void backtrack(List<List<Integer>> list, List<Integer> tmp, int[] nums){
21         if(tmp.size() == nums.length)
22             list.add(new ArrayList<>(tmp));   //要重新new一个list进去
23         else{
24             for(int i = 0; i < nums.length; i++){
25                 if(tmp.contains(nums[i])) continue;
26                 tmp.add(nums[i]);
27                 backtrack(list, tmp, nums);
28                 tmp.remove(tmp.size() - 1); //回溯的重要一步，恢复环境
29 
30             }
31         }
32     }
33 }

47. permutations II

稍微增加了点难度，有重复的数字，采取的技巧其实是相当于对重复的数字人为规定一个顺序

 1 /* Given a collection of numbers that might contain duplicates, return all possible unique permutations.
 2 
 3 For example,
 4 [1,1,2] have the following unique permutations:
 5 [
 6   [1,1,2],
 7   [1,2,1],
 8   [2,1,1]
 9 ] */
10 
11 public class Solution {
12     public List<List<Integer>> permuteUnique(int[] nums) {
13         List<List<Integer>> res = new ArrayList<>();
14         if(nums == null || nums.length == 0) return res;
15         Arrays.sort(nums); //排序不要忘
16         backtrack(res, new ArrayList<>(), nums, new boolean[nums.length]);
17         return res;
18     }
19     public void backtrack(List<List<Integer>> list, List<Integer> tmp, int[] nums, boolean[] used){
20         if(tmp.size() == nums.length)
21             list.add(new ArrayList<>(tmp));
22         else{
23             for(int i = 0; i < nums.length; i++){
24                 if(used[i] || (i > 0 && nums[i-1] == nums[i] && !used[i-1])) continue;
25                 used[i] = true;
26                 tmp.add(nums[i]);
27                 backtrack(list, tmp, nums, used);
28                 used[i] = false;
29                 tmp.remove(tmp.size() - 1);
30             }
31         }
32     }
33 }

posted on 2017-08-13 10:59 MicN 阅读(269) 评论(0) 编辑收藏举报