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LintCode 55 比较字符串(compare string)
阅读量:323 次
发布时间:2019-03-04

本文共 692 字,大约阅读时间需要 2 分钟。

描述

比较两个字符串A和B,确定A中是否包含B中所有的字符。字符串A和B中的字符都是 大写字母

在 A 中出现的 B 字符串里的字符不需要连续或者有序。

您在真实的面试中是否遇到过这个题?

样例
给出 A = “ABCD” B = “ACD”,返回 true

给出 A = “ABCD” B = “AABC”, 返回 false

# 您的提交打败了 85.20% 的提交!class Solution:    """    @param A: A string    @param B: A string    @return: if string A contains all of the characters in B return true else return false    """    def compareStrings(self, A, B):        # write your code here        for i in B:            if i in A:                A = A.replace(i, "", 1)            else:                return False        return True# debug# main = Solution()# reuslt = main.compareStrings("ABCD", "AABC")# print(result)

提交时只需要提交class类内包含的代码,# 后面的代码供在编译器中debug(调试)使用。

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