class Solution: def commonChars(self, A): A.sort(reverse=True) #print(A) x=len(A[0]) l=[] for i in range(len(A)): x=min(x,len(A[i])) #print(x) for j in A[0]: for i in range(1,len(A)): if j not in A[i]: break else: l.append(j) #print(j) for i in range(1,len(A)): #print(list(A[i])) c=list(A[i]) c.remove(str(j)) c=''.join(c) A[i]=c #print(A[i]) return l执行用时 :80 ms, 在所有 Python3 提交中击败了43.61%的用户 内存消耗 :13.9 MB, 在所有 Python3 提交中击败了5.24%的用户 别人48ms的例子:
class Solution: def commonChars(self, A: List[str]) -> List[str]: res=[] if not A: return res key=set(A[0]) for k in key: minnum=min(a.count(k) for a in A) res+=minnum*k return res
我怎么就想不到这样的办法?
minnum=min(a.count(k) for a in A)这个用法好棒!!! ——2019.10.7