LeeCode 14. 最长公共前缀

分治法图示

class Solution:
    # 水平扫描法
    def longestCommonPrefix(self, strs):
        """
        :type strs: List[str]
        :rtype: str
        """
        if len(strs) == 1:
            return strs.first
        reult = ""
        minLength = 10000
        charIndex = 0
        tempChar = ""
        while True:
            flag = False
            for index in range(len(strs)):
                tempStr = strs[index]
                if charIndex == 0:
                    minLength = min(len(tempStr), minLength)
                if index == 0:
                    tempChar = tempStr[charIndex:charIndex+1]
                else:
                    if tempStr[charIndex:charIndex+1] == tempChar:
                        if index+1 == len(strs):
                            reult += tempChar
                        
                    else:
                        flag = True
                        break
            charIndex += 1
            if flag or charIndex > minLength:
                break
        return reult

    # 分治法，空间开销变大了
    # def longestCommonPrefix(self, strs):
    #     l = len(strs)
    #     # print(strs, str(l))
    #     if l > 2:
    #         return self.getCommonPrefix(self.getCommonPrefix(strs[0], strs[1]),self.longestCommonPrefix(strs[2:]))
    #     else:
    #         if l == 1:
    #             return strs[0]
    #         elif l == 2:
    #             return self.getCommonPrefix(strs[0], strs[1])
    #         else:
    #             return ""

    # def getCommonPrefix(self, strs0, strs1):
    #     ret = ""
    #     charIndex = 0
    #     minLen = min(len(strs0), len(strs1))
    #     while charIndex < minLen:
    #         if strs0[charIndex] == strs1[charIndex]:
    #             ret += strs0[charIndex]
    #             charIndex += 1
    #         else:
    #             break
    #     # print(strs0, strs1, ret)
    #     return ret

test = Solution()
# print([0,1,2,3,4,5,6][:3])
# print([0,1,2,3,4,5,6][3:])
# print("123"[2:3])
print(test.longestCommonPrefix(
    ["flower0", "flow1", "flower2", "flow3", "flower4", "flow5", "flight6", "flight7"]))

经过证明，分治法确实要比同步遍历索引慢一些，而且消耗的内存空间更多。