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);
            }
        }
    }
}