47. 全排列 II

输入: [1,1,2]
输出:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

/*
 * @lc app=leetcode id=47 lang=javascript
 *
 * [47] Permutations II
 *
 * https://leetcode.com/problems/permutations-ii/description/
 *
 * algorithms
 * Medium (39.29%)
 * Total Accepted:    234.1K
 * Total Submissions: 586.2K
 * Testcase Example:  '[1,1,2]'
 *
 * Given a collection of numbers that might contain duplicates, return all
 * possible unique permutations.
 *
 * Example:
 *
 *
 * Input: [1,1,2]
 * Output:
 * [
 * ⁠ [1,1,2],
 * ⁠ [1,2,1],
 * ⁠ [2,1,1]
 * ]
 *
 *
 */
function backtrack(list, nums, tempList, visited) {
  if (tempList.length === nums.length) return list.push([...tempList]);
  for (let i = 0; i < nums.length; i++) {
    // 和46.permutations的区别是这道题的nums是可以重复的
    // 我们需要过滤这种情况
    if (visited[i]) continue; // 不能用tempList.includes(nums[i])了，因为有重复
    // visited[i - 1] 这个判断容易忽略
    if (i > 0 && nums[i] === nums[i - 1] && visited[i - 1]) continue;

    visited[i] = true;
    tempList.push(nums[i]);
    backtrack(list, nums, tempList, visited);
    visited[i] = false;
    tempList.pop();
  }
}
/**
 * @param {number[]} nums
 * @return {number[][]}
 */
var permuteUnique = function(nums) {
  const list = [];
  backtrack(list, nums.sort((a, b) => a - b), [], []);
  return list;
};

class Solution:
    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        """与46题一样，当然也可以直接调用itertools的函数，然后去重"""
        return list(set(itertools.permutations(nums)))

    def permuteUnique(self, nums: List[int]) -> List[List[int]]:
        """自己写回溯法，与46题相比，需要去重"""
        # 排序是为了去重
        nums.sort()
        res = []
        def _backtrace(nums, pre_list):
            if len(nums) <= 0:
                res.append(pre_list)
            else:
                for i in range(len(nums)):
                    # 如果是同样的数字，则之前一定已经生成了对应可能
                    if i > 0 and nums[i] == nums[i-1]:
                        continue
                    p_list = pre_list.copy()
                    p_list.append(nums[i])
                    left_nums = nums.copy()
                    left_nums.pop(i)
                    _backtrace(left_nums, p_list)
        _backtrace(nums, [])
        return res