DFS 深度优先遍历

全排列

给定一个 没有重复 数字的序列，返回其所有可能的全排列。

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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]]
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var permute = function(nums) {
  var res = []
  var path = []
  var dfs = (path, nums) => {
    if (path.length === nums.length) {
      res.push(path)
    }
    for (let num of nums) {
      if (!path.includes(num)) {
          path.push(num)
          dfs(path.slice(), nums)
          path.pop()
      }
    }
  }
  dfs(path, nums)
  return res
}