做出了四道题目，但后两道做的很勉强，也错了几次。刚看第三道题，尽管写着 medium，然而没思路，关上第四道看到反而曾经有人过了，于是先做了第四道，才回来做第一道。
最初一道是纯暴力枚举加了几个条件：
- 去除相邻反复值
- 是函数输入为 0 时进行枚举，因为 0 按位与任何数都仍然是 0.
- 因为在一直按位与的过程中函数输入是枯燥减的，所以，但差值大于以后最小差值也能够进行遍历
然而我感觉应该还有更好的解法
第三题用的模仿办法，顺次寻找只含一个字母，两个字母 ... 的串
5464 换酒问题
https://leetcode-cn.com/conte...
class Solution: def numWaterBottles(self, numBottles: int, numExchange: int) -> int: c = numBottles cc = numBottles kp = numBottles while kp >= numExchange: cc = kp // numExchange kp = kp % numExchange + cc c += cc return c5465 子树中标签雷同的节点数
https://leetcode-cn.com/conte...
class Solution: def countSubTrees(self, n: int, edges: List[List[int]], labels: str) -> List[int]: ed = {} for s, e in edges: if s not in ed: ed[s] = [] if e not in ed: ed[e] = [] ed[s].append(e) ed[e].append(s) ans = [0] * n vised = [False] * n def vis(i): vised[i] = True c = {} c[labels[i]] = 1 if i in ed: for ch in ed[i]: if not vised[ch]: cc = vis(ch) for k in cc: if k not in c: c[k] = 0 c[k] += cc[k] ans[i] = c[labels[i]] return c vis(0) return ans5466 最多的不重叠子字符串
https://leetcode-cn.com/conte...
class Solution: def maxNumOfSubstrings(self, s: str) -> List[str]: d = {} for i, ch in enumerate(s): if ch not in d: d[ch] = [i, i, True] else: if d[ch][1] != i - 1: d[ch][2] = False d[ch][1] = i #print(d) used = {} ans = [] nd = [] for ch in d: if d[ch][2]: used[ch] = True ans.append(s[d[ch][0]:d[ch][1]+1]) for ch in d: if not d[ch][2]: cc = 0 ccx = set() f = True mini = d[ch][0] maxi = d[ch][1] change = True while change: change = False i = d[ch][0] while i <= maxi: if s[i] in used: f = False break else: if maxi < d[s[i]][1]: change = True maxi = d[s[i]][1] if mini > d[s[i]][0]: mini = d[s[i]][0] change = True ccx.add(s[i]) i += 1 i = d[ch][0] while i >= mini: if s[i] in used: f = False break else: if maxi < d[s[i]][1]: change = True maxi = d[s[i]][1] if mini > d[s[i]][0]: mini = d[s[i]][0] change = True ccx.add(s[i]) i -= 1 d[ch][1] = maxi d[ch][0] = mini if f: nd.append([len(ccx), maxi-d[ch][0], ch, list(ccx)]) #print(d) nd.sort() for n in nd: f = True for cc in n[3]: if cc in used: f = False break if f: for cc in n[3]: used[cc] = True ch = n[2] ans.append(s[d[ch][0]:d[ch][1]+1]) return ans5467 找到最靠近目标值的函数值
https://leetcode-cn.com/conte...
class Solution: def closestToTarget(self, arr: List[int], target: int) -> int: arr2 = [] l = arr[0] -1 for v in arr: if v != l: arr2.append(v) l = v #print(len(arr), len(arr2)) arr = arr2 r = abs(-1000000000 - target) for i in range(len(arr)): ans = arr[i] r = min(r, abs(ans - target)) for j in range(i+1, len(arr)): ans = ans & arr[j] r = min(r, abs(ans - target)) if ans == 0 or r == 0: break if target > ans and target - ans >= r: break if r == 0: break return r欢送来我的博客刷题分类（有刷题和口试内容）： https://es2q.com/blog/categor...