LeetCode-47. Permutations II

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问题描述

Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order.

Example 1:

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Input: nums = [1,1,2]
Output:
[[1,1,2],
 [1,2,1],
 [2,1,1]]

Example 2:

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Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Constraints:

  • 1 <= nums.length <= 8
  • -10 <= nums[i] <= 10

解答

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/*
 * @Description: 47. Permutations II
 * @Author: libk
 * @Github: https://github.com/libk
 */
const permuteUnique = function (nums) {
  const len = nums.length
  const res = []
  const path = []
  const used = new Array(len).fill(0)

  nums.sort((a, b) => a - b)

  const backTrace = (depth) => {
    if (depth === len) {
      res.push(path.concat())
      return
    }

    for (let i = 0; i < len; i++) {
      // nums[i-1]存在正在path中使用和已经被使用过两种状态，如果已经被
      // 用过，那直接跳过
      if (i > 0 && nums[i] === nums[i - 1] && !used[i - 1]) {
        continue
      }
      if (!used[i]) {
        path.push(nums[i])
        used[i] = 1
        backTrace(depth + 1)
        used[i] = 0
        path.pop(nums[i])
      }
    }
  }

  backTrace(0)

  return res
}