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：

 /* Given a collection of distinct numbers, return all possible permutations.

 For example,
 [1,2,3] have the following permutations:
 [
   [1,2,3],
   [1,3,2],
   [2,1,3],
   [2,3,1],
   [3,1,2],
   [3,2,1]
 ] */

 public class Solution {
     public List<List<Integer>> permute(int[] nums) {
         List<List<Integer>> res = new ArrayList<>();
         backtrack(res, new ArrayList<>(), nums);
         return res;
     }
     public void backtrack(List<List<Integer>> list, List<Integer> tmp, int[] nums){
         if(tmp.size() == nums.length)
             list.add(new ArrayList<>(tmp));   //要重新new一个list进去
         else{
             for(int i = 0; i < nums.length; i++){
                 if(tmp.contains(nums[i])) continue;
                 tmp.add(nums[i]);
                 backtrack(list, tmp, nums);
                 tmp.remove(tmp.size() - 1); //回溯的重要一步，恢复环境

             }
         }
     }
 }

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

 /* Given a collection of numbers that might contain duplicates, return all possible unique permutations.

 For example,
 [1,1,2] have the following unique permutations:
 [
   [1,1,2],
   [1,2,1],
   [2,1,1]
 ] */

 public class Solution {
     public List<List<Integer>> permuteUnique(int[] nums) {
         List<List<Integer>> res = new ArrayList<>();
         if(nums == null || nums.length == 0) return res;
         Arrays.sort(nums); //排序不要忘
         backtrack(res, new ArrayList<>(), nums, new boolean[nums.length]);
         return res;
     }
     public void backtrack(List<List<Integer>> list, List<Integer> tmp, int[] nums, boolean[] used){
         if(tmp.size() == nums.length)
             list.add(new ArrayList<>(tmp));
         else{
             for(int i = 0; i < nums.length; i++){
                 if(used[i] || (i > 0 && nums[i-1] == nums[i] && !used[i-1])) continue;
                 used[i] = true;
                 tmp.add(nums[i]);
                 backtrack(list, tmp, nums, used);
                 used[i] = false;
                 tmp.remove(tmp.size() - 1);
             }
         }
     }
 }