LeetCode题解(0690)：计算员工的重要性(Python)

题目：原题链接（简单）

解法时间复杂度空间复杂度执行用时Ans 1 (Python)$O(N)$$O(N)$168ms (96.22%)Ans 2 (Python)$O(N)$$O(N)$172ms (91.96%)Ans 3 (Python)

LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。

解法一（哈希表）：

def getImportance(self, employees: List['Employee'], id: int) -> int: hashmap = {} for employee in employees: hashmap[employee.id] = employee ans = 0 sub = [id] while len(sub) > 0: k = sub.pop() employee = hashmap[k] ans += employee.importance sub.extend(employee.subordinates) return ans

解法二（递归）：

def getImportance(self, employees: List['Employee'], id: int) -> int: hashmap = {} for employee in employees: hashmap[employee.id] = employee def helper(eid): e = hashmap[eid] return e.importance + sum([helper(s) for s in e.subordinates]) return helper(id)