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666. Path Sum IV 👎

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int pathSum(vector<int>& nums) {
    int ans = 0;
    vector<vector<int>> tree(4, vector<int>(8, -1));

    for (const int num : nums) {
      const int d = num / 100 - 1;
      const int p = (num % 100) / 10 - 1;
      const int v = num % 10;
      tree[d][p] = v;
    }

    dfs(tree, 0, 0, 0, ans);
    return ans;
  }

 private:
  void dfs(const vector<vector<int>>& tree, int i, int j, int path, int& ans) {
    if (tree[i][j] == -1)
      return;
    if (i == 3 || max(tree[i + 1][j * 2], tree[i + 1][j * 2 + 1]) == -1) {
      ans += path + tree[i][j];
      return;
    }

    dfs(tree, i + 1, j * 2, path + tree[i][j], ans);
    dfs(tree, i + 1, j * 2 + 1, path + tree[i][j], ans);
  }
};
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class Solution {
  public int pathSum(int[] nums) {
    int[][] tree = new int[4][8];
    Arrays.stream(tree).forEach(A -> Arrays.fill(A, -1));

    for (final int num : nums) {
      final int d = num / 100 - 1;
      final int p = (num % 100) / 10 - 1;
      final int v = num % 10;
      tree[d][p] = v;
    }

    dfs(tree, 0, 0, 0);
    return ans;
  }

  private int ans = 0;

  private void dfs(int[][] tree, int i, int j, int path) {
    if (tree[i][j] == -1)
      return;
    if (i == 3 || Math.max(tree[i + 1][j * 2], tree[i + 1][j * 2 + 1]) == -1) {
      ans += path + tree[i][j];
      return;
    }

    dfs(tree, i + 1, j * 2, path + tree[i][j]);
    dfs(tree, i + 1, j * 2 + 1, path + tree[i][j]);
  }
}
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class Solution:
  def pathSum(self, nums: List[int]) -> int:
    ans = 0
    tree = [[-1] * 8 for _ in range(4)]

    for num in nums:
      d = num // 100 - 1
      p = (num % 100) // 10 - 1
      v = num % 10
      tree[d][p] = v

    def dfs(i: int, j: int, path: int) -> None:
      nonlocal ans
      if tree[i][j] == -1:
        return
      if i == 3 or max(tree[i + 1][j * 2], tree[i + 1][j * 2 + 1]) == -1:
        ans += path + tree[i][j]
        return

      dfs(i + 1, j * 2, path + tree[i][j])
      dfs(i + 1, j * 2 + 1, path + tree[i][j])

    dfs(0, 0, 0)
    return ans