打算随意的做几题，看到 50 号问题几百个字的描述我却使用一行代码仅仅 11 个字母就 AC，哎，枯燥。
for v in g[u]: if v not in states: if not dfs(v): return False elif states[v] == VISITING: return False order.append(u) states[u] = VISITED return True return ''.join(reversed(order)) if all(dfs(u) for u in g if u not in states) else "" Java class Solution { static ...
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请注意,在使用完结果字符串后,不要忘记使用free()函数释放分配的内存。 Python3版本 def longestCommonPrefix(strs): # 如果字符串数组为空或长度为0,直接返回空字符串 if not strs: return "" # 找出最短字符串的长度 minLength = min(len(s) for s in strs) # 使用二分法查找最长公共前缀 low = 1...
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Course Schedule, Longest Increasing Path in a Matrix and Alien Dictionary. Note that some of the tree problems can also be asked in n-ary tree format, so make sure you know what an n-ary tree is. Note: The last problem may be quite tricky to solve, so feel free to skip that one....
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