Leetcode 算法面试冲刺 热题 HOT 100 刷题（337 338 347 394 399）（六十八）

337. 打家劫舍 III

直接上答案：

看了b站一个小姐姐的讲解，写出来的：
https://www.bilibili.com/video/BV1XL411w7oM/?spm_id_from=333.788

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

# [0, 0]
# [0]代表不偷，[1]代表偷
# root 不偷，那么就取left的[0][1]中的最大值 + right[0][1]中的最大值
# root 偷，root.val + left[0] + right[0]
class Solution:
    def rob(self, root: Optional[TreeNode]) -> int:
        def helper(root):
            if not root: return [0, 0]
            l = helper(root.left)
            r = helper(root.right)
            no_rob = max(l) + max(r)
            rob = root.val + l[0] + r[0]
            return [no_rob, rob]
        return max(helper(root))

338. 比特位计数

看了如何计算二进制的1，就是和自己的i-1的数字去&，一直&到i=0.

class Solution:
    def countBits(self, n: int) -> List[int]:
        ans = [0]
        for i in range(1, n + 1):
            count = 0
            while i != 0:
                i &= i - 1
                count += 1
            ans.append(count)
        return ans
                

347. 前 K 个高频元素

看了答案知道是小顶堆，然后自己写出来的：

import heapq
class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        dic = {}
        for num in nums:
            if num in dic:
                dic[num] += 1
            else:
                dic[num] = 1

        heap = []
        for num, cnt in dic.items():
            if len(heap) < k:
                heapq.heappush(heap, [cnt, num])
            else:
                tmp = heapq.heappop(heap)
                if cnt > tmp[0]:
                    heapq.heappush(heap, [cnt, num])
                else:
                    heapq.heappush(heap, tmp)

        return [i[1] for i in heap]

394. 字符串解码

想了用栈的方法，但是觉得过程有点复杂，看了K神答案：










K神代码

class Solution:
    def decodeString(self, s: str) -> str:
        stack, res, multi = [], "", 0
        for c in s:
            if c == '[':
                stack.append([multi, res])
                res, multi = "", 0
            elif c == ']':
                cur_multi, last_res = stack.pop()
                res = last_res + cur_multi * res
            elif '0' <= c <= '9':
                multi = multi * 10 + int(c)            
            else:
                res += c
        return res

按照K神思路，写的，基本一样，命名比我的好：

class Solution:
    def decodeString(self, s: str) -> str:
        stack, res, multi = [], "", 0
        for c in s:
            if c == "[":
                stack.append([multi, res])
                res, multi = "", 0
            elif c == "]":
                tmp_multi, tmp_r = stack.pop()
                res = tmp_r + tmp_multi * res

            elif '0' <= c <= '9':
                multi = multi * 10 + int(c)
            else:
                res += c
        return res

399. 除法求值

不会。先跳吧，答案用了并查集，很难。