LeetCode 6. ZigZag Conversion & 字符串

ZigZag Conversion

看了三遍题目才懂，都有点怀疑自己是不是够聪明...

就是排成这个样子啦，然后从左往右逐行读取返回。
这题看起来很简单，做起来，应该也很简单。

通过位置计算行数：

P     I    N
A   L S  I G
Y A   H R
P     I
0,0 1,1 2,2 3,3 4,2 5,1 6,0 (期望)

用简单的先思考：

P   A   H   N
A P L S I I G
Y   I   R
0,0 1,1 2,2 3,3 4,0       %4 (3+1)可以解决正常的
            3,1(期望)     特殊的3(2行2排的P，用1- (%4-row) )

这里(3+1)以及1- (%4-row)中的1作为取余底数的延长

找某个键所在行数

# find k in line
def findLine(k,rows):

    # 延长，去掉斜路两头
    prolong=rows-2 
    
    # 取模底
    modulo=rows+prolong
    print(modulo,prolong+1,k%modulo,(k%modulo-rows+1)%(prolong+1),(k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows))
    return (k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows+1)%(prolong+1)

第1次提交

class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """
        zigzag=''        

        # each line add 'c' to zigzag string.
        for n in range(numRows):
            for k,c in enumerate(s):
                #print(k,c,end=" ")
                if findLine(k,numRows) == n:
                    zigzag=zigzag+c  

        return zigzag
# find k in line
def findLine(k,rows):

    # 延长，去掉斜路两头
    prolong=rows-2 
    
    # 取模底
    modulo=rows+prolong
    #print(modulo,prolong+1,k%modulo,(k%modulo-rows+1)%(prolong+1),(k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows))
    return (k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows)


if __name__ == "__main__":
    
    data = [
        {
            "input":{
                's':'PAYPALISHIRING',
                'numRows':4
            },
            "output":"PINALSIGYAHRPI", 
        },{
            "input":{
                's':'PAYPALISHIRING',
                'numRows':3
            },
            "output":"PAHNAPLSIIGYIR", 
        },

    ];
    for d in data:
        print(d['input']['s'],d['input']['numRows'])
        result=Solution().convert(d['input']['s'],d['input']['numRows'])
        print(result)
        if result==d['output']:
            print("--- ok ---")
        else:
            print("--- error ---")

Runtime Error:

Runtime Error Message:
Line 27: ZeroDivisionError: integer division or modulo by zero
Last executed input:
"A"
1

粗心+1，findLine()函数里没有检验值的有效性。

第2次提交

# find k in line
def findLine(k,rows):
    # 延长，去掉斜路两头
    prolong=rows-2 
    
    # 取模底
    modulo=rows+prolong
    
    # validity check
    if prolong<0:
        prolong=0
    if modulo<=0:
        modulo=1

    #print(modulo,prolong+1,k%modulo,(k%modulo-rows+1)%(prolong+1),(k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows))
    return (k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows)

Time Limit Exceeded:

"kvzeeubynglxfdedshtpobqsdhufkzgwuhaabdzrlkosnuxibrxssnkxuhcggkecshdvkcmymdqbxolbfjtzyfwtmbbungzfpcbbgpzusqxqejrlsmkqtglijpcxxbcmffnlvnfpddfjmyugkeyemkmyzqvwszkxfxlckqrpvzyjxupkyoonaclbsgzmhjmogxstpkilljwidoseeitemefhmgtvpfkxecquobhzkfkptetxpmdbskigqecflmdqqvmfwveiaqyuvrtkgxlyhwhyalfnzifpgrucoblprjloceykbkjlisjkdoxczdtfwqjlrwckhnzkrxuvjfgtzrdchdgiicneszrlvtxdiwncwjxhrfbqygvfjdorfdyzcrkylidvgqxebwmubplzxihjlvataasdsfdfngavyyabuowyfhzcpglcdoxeoqjivmnkuofsohtivpiayifpoquugryvjjfgvtqrjyjxhefdwqfwykmodiijzigjrmpohifqiqnpvuutkcpiodzrljdlslwlxnagxhwfylxvgtosvfdkjcdulihfudrtrtaoaywakvvqolkmtnycpdwdmeigjbbcubrxapxmkveaiombckftocwaifitgjwdnpapezbqwhqhvdizpotdspfcwpxfbtiqikfolieipxpmazmrphxjyenvulcxeknpwsfhckptjgflitczczjbeyyajaxqmkhiempgyfzhngsvcvxewghcgfcqhzitlpbpbrvaywjlfcjhzgnxoxauecmmeufpljfpacrazaneewndecbuzbrgffsjczznieckitkhwynawcgdfjzgmqmrygbaicpqiudqpnylnnoksupzdofphuifcjhknydvsgmivmvjbjttdksiyazhuimytvjhuocmuqwpcsyedtzjdsresrlozamsvxbrlegfucxzwxfcrelwyeaqvoewotrlssdeyjltnkumibozfzxe"
200

超时了，加一个时钟本地看看时间：

运行时间

PAYPALISHIRING 4
PINALSIGYAHRPI
--- ok ---      -3.590000000000017e-05
PAYPALISHIRING 3
PAHNAPLSIIGYIR
--- ok ---      -2.6699999999999294e-05
A 1
A
--- ok ---      -8.399999999998686e-06
kvzeeubynglxfdedshtpobqsdhufkzgwuhaabdzrlkosnuxibrxssnkxuhcggkecshdvkcmymdqbxolbfjtzyfwtmbbungzfpcbbgpzusqxqejrlsmkqtglijpcxxbcmffnlvnfpddfjmyugkeyemkmyzqvwszkxfxlckqrpvzyjxupkyoonaclbsgzmhjmogxstpkilljwidoseeitemefhmgtvpfkxecquobhzkfkptetxpmdbskigqecflmdqqvmfwveiaqyuvrtkgxlyhwhyalfnzifpgrucoblprjloceykbkjlisjkdoxczdtfwqjlrwckhnzkrxuvjfgtzrdchdgiicneszrlvtxdiwncwjxhrfbqygvfjdorfdyzcrkylidvgqxebwmubplzxihjlvataasdsfdfngavyyabuowyfhzcpglcdoxeoqjivmnkuofsohtivpiayifpoquugryvjjfgvtqrjyjxhefdwqfwykmodiijzigjrmpohifqiqnpvuutkcpiodzrljdlslwlxnagxhwfylxvgtosvfdkjcdulihfudrtrtaoaywakvvqolkmtnycpdwdmeigjbbcubrxapxmkveaiombckftocwaifitgjwdnpapezbqwhqhvdizpotdspfcwpxfbtiqikfolieipxpmazmrphxjyenvulcxeknpwsfhckptjgflitczczjbeyyajaxqmkhiempgyfzhngsvcvxewghcgfcqhzitlpbpbrvaywjlfcjhzgnxoxauecmmeufpljfpacrazaneewndecbuzbrgffsjczznieckitkhwynawcgdfjzgmqmrygbaicpqiudqpnylnnoksupzdofphuifcjhknydvsgmivmvjbjttdksiyazhuimytvjhuocmuqwpcsyedtzjdsresrlozamsvxbrlegfucxzwxfcrelwyeaqvoewotrlssdeyjltnkumibozfzxe 200
kmavwupczbbfreepjaexllzuqzpabgxfnyviuendheegijmwlllmnxyvcdfkaeedrtucecaabdzaxusysozhddxbtfsnrprfggoodzfbdfhfqjnjssfgcjdvafchgvlzuyyjzfqywnkbayizfbaegruvcwhorkuxwbihjyptawfbkachphbnzlwdwctyzipinrdgzalxlhwktcqcovdcgslofdnrxgfuzecjxsohzieqggbnjwmrcieqxivxmsimvrsgncyndkvgkhusbxcogaudfnihrshcczozpgthfqggtyikfiguejvpdcvpmqsuiephxaindryhyvkiklkzfmncnpqnmhoxoykqakmcujsdwuauqrgypblryzxjyedoqvbolwjjfbfjzpftfchjdgzutzvcizcttfyxqicforljwdjfhtkygkmjjjnbsxtybihpdulekvnjfcsgkdhgzbwfmfkqsipyfwvcewpmbcynvbokkjglmebpjoxjzrdctupiltsliudqbjvkxoznsqcieieugyyjrjjargrxzlpmhhsfppumiorikzhmmqnizytffatglqmvlaipjiyqxhjhnpupwpiochvecxyuimxlulubxtoqcgkfwmkckpftpicfriqsnvoiyludtevyzbdnqrftfalxzpijpjdedwddvlcsfwsfrjflpemmwssyvldruqxtlgqnookdapzemgzaylximefhdsmcwvvkefhxmqyqbyglhrzixwlqkvqevsgbgwbtzfsdoeuzmspckpvaxxxfpzftdnwxekdxltjwfcpcjckkdgrqfuterklilpzifwvhhiyzbfaeyouwajudcqxqrovucttoperfekxtkwykacoofobtopamrnvyolatwiscgaaslmkedbhvvesfvkygeqmjzmoxlmelpthtkanjimxkmetruoenbmgsyuixoccbsdpbotidbzpwwjfkjdgzilmixlee
--- ok ---      -0.07744500000000001

比其他的是慢了1000倍，0.77ms，目标：优化到1ms.

分析

问题就在于，外层的(计算层数的)for没必要循环的（随着层数增大，每一层都O(n)，多少无用功），只要分numRows个列表，一个for就遍历完。

改完测一下时间

PAYPALISHIRING 4
PINALSIGYAHRPI
--- ok ---      -2.4100000000000857e-05
PAYPALISHIRING 3
PAHNAPLSIIGYIR
--- ok ---      -1.9699999999999232e-05
A 1
A
--- ok ---      -1.1399999999999952e-05
kvzeeubynglxfdedshtpobqsdhufkzgwuhaabdzrlkosnuxibrxssnkxuhcggkecshdvkcmymdqbxolbfjtzyfwtmbbungzfpcbbgpzusqxqejrlsmkqtglijpcxxbcmffnlvnfpddfjmyugkeyemkmyzqvwszkxfxlckqrpvzyjxupkyoonaclbsgzmhjmogxstpkilljwidoseeitemefhmgtvpfkxecquobhzkfkptetxpmdbskigqecflmdqqvmfwveiaqyuvrtkgxlyhwhyalfnzifpgrucoblprjloceykbkjlisjkdoxczdtfwqjlrwckhnzkrxuvjfgtzrdchdgiicneszrlvtxdiwncwjxhrfbqygvfjdorfdyzcrkylidvgqxebwmubplzxihjlvataasdsfdfngavyyabuowyfhzcpglcdoxeoqjivmnkuofsohtivpiayifpoquugryvjjfgvtqrjyjxhefdwqfwykmodiijzigjrmpohifqiqnpvuutkcpiodzrljdlslwlxnagxhwfylxvgtosvfdkjcdulihfudrtrtaoaywakvvqolkmtnycpdwdmeigjbbcubrxapxmkveaiombckftocwaifitgjwdnpapezbqwhqhvdizpotdspfcwpxfbtiqikfolieipxpmazmrphxjyenvulcxeknpwsfhckptjgflitczczjbeyyajaxqmkhiempgyfzhngsvcvxewghcgfcqhzitlpbpbrvaywjlfcjhzgnxoxauecmmeufpljfpacrazaneewndecbuzbrgffsjczznieckitkhwynawcgdfjzgmqmrygbaicpqiudqpnylnnoksupzdofphuifcjhknydvsgmivmvjbjttdksiyazhuimytvjhuocmuqwpcsyedtzjdsresrlozamsvxbrlegfucxzwxfcrelwyeaqvoewotrlssdeyjltnkumibozfzxe 200
kmavwupczbbfreepjaexllzuqzpabgxfnyviuendheegijmwlllmnxyvcdfkaeedrtucecaabdzaxusysozhddxbtfsnrprfggoodzfbdfhfqjnjssfgcjdvafchgvlzuyyjzfqywnkbayizfbaegruvcwhorkuxwbihjyptawfbkachphbnzlwdwctyzipinrdgzalxlhwktcqcovdcgslofdnrxgfuzecjxsohzieqggbnjwmrcieqxivxmsimvrsgncyndkvgkhusbxcogaudfnihrshcczozpgthfqggtyikfiguejvpdcvpmqsuiephxaindryhyvkiklkzfmncnpqnmhoxoykqakmcujsdwuauqrgypblryzxjyedoqvbolwjjfbfjzpftfchjdgzutzvcizcttfyxqicforljwdjfhtkygkmjjjnbsxtybihpdulekvnjfcsgkdhgzbwfmfkqsipyfwvcewpmbcynvbokkjglmebpjoxjzrdctupiltsliudqbjvkxoznsqcieieugyyjrjjargrxzlpmhhsfppumiorikzhmmqnizytffatglqmvlaipjiyqxhjhnpupwpiochvecxyuimxlulubxtoqcgkfwmkckpftpicfriqsnvoiyludtevyzbdnqrftfalxzpijpjdedwddvlcsfwsfrjflpemmwssyvldruqxtlgqnookdapzemgzaylximefhdsmcwvvkefhxmqyqbyglhrzixwlqkvqevsgbgwbtzfsdoeuzmspckpvaxxxfpzftdnwxekdxltjwfcpcjckkdgrqfuterklilpzifwvhhiyzbfaeyouwajudcqxqrovucttoperfekxtkwykacoofobtopamrnvyolatwiscgaaslmkedbhvvesfvkygeqmjzmoxlmelpthtkanjimxkmetruoenbmgsyuixoccbsdpbotidbzpwwjfkjdgzilmixlee
--- ok ---      -0.0005619000000000006

bingo! 0.56ms
提交看看，肯定AC

第3次提交 完整代码

import time

class Solution:
    def convert(self, s, numRows):
        """
        :type s: str
        :type numRows: int
        :rtype: str
        """       
        # create list array save each line
        listStr=[]
        for n in range(numRows):
            try:
                listStr[n]
            except:
                listStr.append('')

        # each line add char to listStr[n].
        for k,c in enumerate(s):
            #print(k,c,end=" ")
            n=findLine(k,numRows)
            listStr[n]=listStr[n]+c

        # joint str
        zigzag=''.join(listStr)
        return zigzag
# find k in line
def findLine(k,rows):
    # 延长，去掉斜路两头
    prolong=rows-2 
    
    # 取模底
    modulo=rows+prolong

    # validity check
    if prolong<0:
        prolong=0
    if modulo<=0:
        modulo=1

    #print(modulo,prolong+1,k%modulo,(k%modulo-rows+1)%(prolong+1),(k%modulo) if (k%modulo)<rows else prolong-(k%modulo-rows))
    x=k%modulo
    return (x) if (x)<rows else prolong-(x-rows)


if __name__ == "__main__":
    
    data = [
        {
            "input":{
                's':'PAYPALISHIRING',
                'numRows':4
            },
            "output":"PINALSIGYAHRPI", 
        },{
            "input":{
                's':'PAYPALISHIRING',
                'numRows':3
            },
            "output":"PAHNAPLSIIGYIR", 
        },
        {
            "input":{
                's':'A',
                'numRows':1
            },
            "output":"A", 
        },
        {
            "input":{
                's':'kvzeeubynglxfdedshtpobqsdhufkzgwuhaabdzrlkosnuxibrxssnkxuhcggkecshdvkcmymdqbxolbfjtzyfwtmbbungzfpcbbgpzusqxqejrlsmkqtglijpcxxbcmffnlvnfpddfjmyugkeyemkmyzqvwszkxfxlckqrpvzyjxupkyoonaclbsgzmhjmogxstpkilljwidoseeitemefhmgtvpfkxecquobhzkfkptetxpmdbskigqecflmdqqvmfwveiaqyuvrtkgxlyhwhyalfnzifpgrucoblprjloceykbkjlisjkdoxczdtfwqjlrwckhnzkrxuvjfgtzrdchdgiicneszrlvtxdiwncwjxhrfbqygvfjdorfdyzcrkylidvgqxebwmubplzxihjlvataasdsfdfngavyyabuowyfhzcpglcdoxeoqjivmnkuofsohtivpiayifpoquugryvjjfgvtqrjyjxhefdwqfwykmodiijzigjrmpohifqiqnpvuutkcpiodzrljdlslwlxnagxhwfylxvgtosvfdkjcdulihfudrtrtaoaywakvvqolkmtnycpdwdmeigjbbcubrxapxmkveaiombckftocwaifitgjwdnpapezbqwhqhvdizpotdspfcwpxfbtiqikfolieipxpmazmrphxjyenvulcxeknpwsfhckptjgflitczczjbeyyajaxqmkhiempgyfzhngsvcvxewghcgfcqhzitlpbpbrvaywjlfcjhzgnxoxauecmmeufpljfpacrazaneewndecbuzbrgffsjczznieckitkhwynawcgdfjzgmqmrygbaicpqiudqpnylnnoksupzdofphuifcjhknydvsgmivmvjbjttdksiyazhuimytvjhuocmuqwpcsyedtzjdsresrlozamsvxbrlegfucxzwxfcrelwyeaqvoewotrlssdeyjltnkumibozfzxe',
                'numRows':200
            },
            "output":"kmavwupczbbfreepjaexllzuqzpabgxfnyviuendheegijmwlllmnxyvcdfkaeedrtucecaabdzaxusysozhddxbtfsnrprfggoodzfbdfhfqjnjssfgcjdvafchgvlzuyyjzfqywnkbayizfbaegruvcwhorkuxwbihjyptawfbkachphbnzlwdwctyzipinrdgzalxlhwktcqcovdcgslofdnrxgfuzecjxsohzieqggbnjwmrcieqxivxmsimvrsgncyndkvgkhusbxcogaudfnihrshcczozpgthfqggtyikfiguejvpdcvpmqsuiephxaindryhyvkiklkzfmncnpqnmhoxoykqakmcujsdwuauqrgypblryzxjyedoqvbolwjjfbfjzpftfchjdgzutzvcizcttfyxqicforljwdjfhtkygkmjjjnbsxtybihpdulekvnjfcsgkdhgzbwfmfkqsipyfwvcewpmbcynvbokkjglmebpjoxjzrdctupiltsliudqbjvkxoznsqcieieugyyjrjjargrxzlpmhhsfppumiorikzhmmqnizytffatglqmvlaipjiyqxhjhnpupwpiochvecxyuimxlulubxtoqcgkfwmkckpftpicfriqsnvoiyludtevyzbdnqrftfalxzpijpjdedwddvlcsfwsfrjflpemmwssyvldruqxtlgqnookdapzemgzaylximefhdsmcwvvkefhxmqyqbyglhrzixwlqkvqevsgbgwbtzfsdoeuzmspckpvaxxxfpzftdnwxekdxltjwfcpcjckkdgrqfuterklilpzifwvhhiyzbfaeyouwajudcqxqrovucttoperfekxtkwykacoofobtopamrnvyolatwiscgaaslmkedbhvvesfvkygeqmjzmoxlmelpthtkanjimxkmetruoenbmgsyuixoccbsdpbotidbzpwwjfkjdgzilmixlee", 
        },

    ];
    for d in data:
        
        print(d['input']['s'],d['input']['numRows'])
        
        # 计算运行时间
        start = time.perf_counter()
        result=Solution().convert(d['input']['s'],d['input']['numRows'])
        end = time.perf_counter()
        
        print(result)
        if result==d['output']:
            print("--- ok ---",end="	")
        else:
            print("--- error ---",end="	")
        
        print(start-end)

总结：好多坑没有踩过就不会有经验，这就是为什么要多码多思考了，还需要一直提升一直充电。不总结，就不会有什么沉淀；不提升，就不会用更高级的方法去建模，去看待这个魔幻的世界。