LeetCode318：单词长度的最大乘积

难度：medium
本题同时也是《剑指offer 专项突击》的第5题【整数专题】。

problem

https://leetcode.cn/problems/aseY1I/
or
https://leetcode-cn.com/problems/maximum-product-of-word-lengths/

给定一个字符串数组 words，请计算当两个字符串 words[i] 和 words[j] 不包含相同字符时，它们长度的乘积的最大值。假设字符串中只包含英语的小写字母。如果没有不包含相同字符的一对字符串，返回 0。

初步解法

这个问题的关键在于判断两个字符串有没有相同的字符。最直观的是扫描str1中的每个字符是否存在于str2中。这里不再赘述。

用哈希表记录字符串中出现的字符

简单来说就是建立一个26*n的表，n表示words中单词个数。每个坐标的值为0 或1 ，代表第n个单词中有没有出现 ‘a’,‘b’,‘c’…。具体做法就是建立一个字典，key为26个字母字符， value为列表，eg.，[0,1,1,0,1,0] 这种。
代码：

class Solution:
    def maxProduct(self, words) -> int:
        t1 = time.time()
        dictw = dict()
        for k in range(ord('a'),ord('z')+1):
            dictw[chr(k)] = []
            for wl in range(len(words)):
                if chr(k) in words[wl]:
                    dictw[chr(k)].append(1)
                else:
                    dictw[chr(k)].append(0)
        num_max = 0
        for ii in range(len(words)-1):
            for jj in range(ii+1, len(words)):
                flag = 0
                for k in dictw.keys():
                    if dictw[k][ii] and dictw[k][jj]:
                        flag+=1
                if flag==0:
                    num_max = max(num_max, len(words[ii])*len(words[jj]))

测试用例：

words =["gfgc","bfdcabbbdbfd","acbeaacfecfbdagaebbeagddgb","fdfefbgffegfcc","cagcbgg","eff","fcdabfcaa","gggdcfacfcdeecfdcbfaeadbec","eaccfagbcagafagfebgdc","aaebefcebcfe","ebdcge","aaeecgcacaece","gfgdafecedbgbaccggegbadfcca","bebefebffdfabcfdg","fdcdaggdcaceabgadgbgab","agdcggbcdagbbcbgaeagaggf","caccecbaggfa","ecegegdcaef","ffaadgaedggabaccebggc","acgaeeafffbbebfcdfdgebcbbg","aacdefgfbaedggfe","bbcfgddgeafcfdbadccgade","bddeacgaggfeeageaaagaccdga","ge","gbgbgebefgefddfdfg","adfdeegacaeabafeafeggdbdabe","ggafeacfgeafggfefggb","afggdcacfafbdcgbb","agedabcfb","dgdbfa","cccgcdfedafgbccdbagebe","dgaadebcggbagbafgbaaecd","gg","fggbeeefdcgcfaeacbbadbfgd","abdegggfeacfcacgceadgbfddc","febd","afdabadgeedbcg","ggc","acefcbec","eaeffebcfgeffccafdb","aeedfc","eabcd","cfafcecbdegd","bgegdccdcgebbecegbgg","bdadfdgdceegd","gefecgbacedcafcgeec","bdfbbac","debgcggbggbaeacgaddcbbefcde","fccfe","acfacgaegeg","cdcfecadbfcag","efc","caccdaafegbdaafedbabfebfe","dcbedbaagbccd","gcbgceaefgfbceegf","ab","aee","aaa","gecaccebbbdgd","eeaafgcggedbgcfadcedffgb","abcbbgddfbecceebg","fd","bccgcgagcdd","ccagbdfagdecgdc","fcbfedaagde","cgcebebecfgffgcdgc","cbeedgdgecbfcfdcgbdbcd","cagccdedbffcfabcacegbe","bgfbebdbbecabgagafagcecdabf","bcddffgefeaabbddedgbee","abbfade","agggbebegcegebfe","baaecbgbdaaaged","fad","daccccdfdgedbdgebcaeeecaba","cfeffaegg","bcaac","aadbcbddedbbabf","bdcfegccfgefb","aab","fggagcf","be","eabf","cbg","cbggbededceffddabbfcg","aegdfffdbc","ffggdgga","aeb","gfbdeegadccdbgedcebce","ebcgdbc","aad","fegeeeeaccfbfdgdefgef","fbgaaaddcf","ceebfdegcegdgdcbbddadfe","egdbfabgfafgaeafagb","cbdg","ebceedadcfd","cbbgecfeegbadbea","cgccbfbgaaeceff","ebcffgbfdggfbbgdeeaddbdec","dbcbddgdcegbabbgfgecdbgfe","cddfgcfbabdffaadfgeebdgd","cfcaaafaab","cegdaafcffbfedgaeeedbecb","fdaeggagcaecadadaacgfdg","fafagffbegcedededaaeaf","bbdabacabcfgage","edeaddebgagacdagecebebge","eeecfg","eefbcefdcabadecgfcfdfgefa","dgccagdaed","bbegfdfeacaddgacddbbfgcfde","bdc","affgecdbgeebffdfadg","adcafc","cbagc","cbbffebea","agfa","efbcffegabfeggdegdd","eededbdegdefeebfebbaf","agacdagfebdadaddedcddg","bafadccdfdbcfbbdgagafg","ebaefgfg","ecgbacffdfgdagae","gfedefgecbcffagbagbeeefaeab","dabebafabcfebbebbda","agdgedccdccdgcbgdcebfbaccf","fagacdccbgagabebac","efaeb","bbfadbbgdcgbefdfbeededabd","gc","ddebgedfgcfeedgdbagebfceafe","gadeabb","cecgeagddcdf","gbccdgccccgbdcbebabebfba","deegcee","fcedcabeg","gdfdgebggcadaeafcddbeegfc","agcfdbdg","bbafdbdccdcfgfe","fcbcdee","ebeddcagadccbadd","afadebbefcddgdg","acfdeedcgdcecfagacdacg","bcdbcgegegdgcdcgcfcggafda","acdcfaacedegebccc","addabdcgegdcdbbaeabfggg","fcedefdeeaeaeab","befdggaff","eafggc","dcdccadcdfebca","eedecdbcgfeabggbge","edccadbgeggb","bbagdcfbaac","fegfddaeedfd","abdgeaefbefdfdgb","eadegdd","cdcffaadfcaedeeegacc","bafaebcgaca","feeggdbgfeecgfefabccg","fegaddf","bfgffgffdffcffagbdc","bgcbcdbdgcfga","ddegbbfa","ddedccefecdgfdaccdadfffbaeb","cfgbfaaabbbbgaeabadccdeeff","fddgdbeabfggafdgcbede","gggfffgdgdgdfdcaefegd","fdc","fffcgfeadadgefcgaefccff","gggebebfgggfeeafbbdefb","afedacbeffefcfddbfdaefdfd","bbdffgacfgaebcab","bcfeggaecbeafcddgbbgce","dfgffbeegbba","gadbf","egcbgedgda","fbfbefce","cfdcacfcdgbaagag","egaae","edbabfcfcddgfffbggecfdbfcfg","ef","bfagffb","adgcdbaafagd","dfgeeddggaebabececbbeba","aabccfgggafgbebac","acgc","dcfbdbgfg","fcbgacfbde","fabdadadgfc","dggfegee","dgggaefeddaeafggaa","geegfabeeadddgbgegagdcecbad","ffbeacbegecegdaadcc","facbfdeeafddececf","fddfebfd","efb","egebdfggdccdecdcdagfgacdcc","eeebbdbefcdfacdbc","aaedbedfaegadfagec","agcddbab","ffabffgdgcdbeeacgagfae","ggfea","begdg","ceddafddgb","bafgbddabefbgedagadgeadg","gbcfbgf","gagfggdbfccc","cffag","cccegcfagdeefggbdcf","adcdgac","ffbg","ccgcedecceb","adegacd","egbbgfacg","cabece","bcaeaecddbefbdf","edbedbag","eaffdgfgcdbdfdccagaad","fdfgebddgbebgbbbdffge","eadaccgbaf","aedfeccdffeffabgdfbea","bdceebcgaccedbdabffdffbfag","fdgfegeeabeegbcaebdf","cdebccda","ddabfdfbefbeaecbcabcdbd","fdedeeadcbedfdbdfgcf","cfgabaefcbdfcbfdbbcafdcdeb","fecccacgcecgcbbadabbcace","caadeeadbfgebdbaa","bcbgbffcbbbfbf","bfgdfeac","fceccaagaa","bagbcfefbeddgec","cdfbbbbadeacacbgaffccfbegg","adcageadebeecgdeefegfaaadde","daabfggdeafebeaad","addaefbefbdfgegdbcffgdedefe","ddadbbcccfbacaceffde","bggeagfedg","bfgabcbbcdbgdgg","cefeagbcfae","ceafefb","gcbdegfaeaegbc","dcgcefdcgfggeefbeacebfg","ccdfce","faccebdgcbegeedccfeececbc","efabeaacfddacadfgfcgd","addfaeaaafffd","agc","edfdgcabecffacefdcecfbfegc","ddaeaeaeccedgbccfefbee","ag","cedfcacaagadfcgddcgb","dfagcceccgdbbcdc","gbceee","edfdaabeedbfea","cgbcaeaebafcbccbcdbecgcdbg","ffadceefbfeccebf","gf","edacdffff","egbf","fefgfeecfgfdag","afeaaebcbcacgcgdde","cgffbafeggggdbeebabg","bbgggfadbecegdaafcg","db","bddgaaaacbfgagfd","becfaabeedcgg","bagggagfdcggdgfbfc","fbafcefbefbcaeceabdegdef","gfccggfgefd","cfdebac","dfbcfecdgfgfbggdggaabcee","acddabagddggaeefdbbcfebbfd","gabcceafdcaeefbae","fegcecgeggbdgcbcdg","baffcegadcfaffbfcbedaecaf","dbdaedcgcgbcgcacafabgca","dgbeccbbbcdegffe","eeadefede","cedbgfgeabcgagebe","ffdceegag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时间：0.2157015800476074

优化

来自：老汤 简单易懂Java/C++ /Python/js/go - 最大单词长度乘积
简单来说，就是建立这样一个字典：{表示abcd出现情况的26位二进制整数 : 单词长度}， 比如对单词 ‘abcd’ 就是 { [1111 00000000…] : 4}，为了表示方便，可以把前面的这个列表转为十进制整数作为键值。同时对相同键值的，直接取最大单词长度更新。
把原博客比较复杂的位操作搞成了列表。

完整代码

import time
import math
class Solution:
    def maxProduct(self, words) -> int:
        t1= time.time()
        bitmask_map, ans = {
    }, 0
        for k in range(len(words)):
            length = len(words[k])
            bitmask =[]
            for abc in range(ord('a'), ord('z')+1):
                abc = chr(abc)
                # 循环'a' 到'z'，看看 abc 是否存在于第 k 个词中, 以26位二进制码形式存储
                if abc in words[k]:
                    bitmask.append(1)
                else:
                    bitmask.append(0)
            # 26位二进制码 转为 十进制整数, 作为字典键值
            bitmask = int(''.join(list(map(str,bitmask))), 2)
            # 这里是合并一下 abc 出现一样的词
            if bitmask in bitmask_map:
                bitmask_map[bitmask] = max(length, bitmask_map[bitmask] )
            else:
                bitmask_map[bitmask] = length
        
        # 字典里的两两匹配 算 长度乘积最大值
        for x in bitmask_map:
            for y in bitmask_map:
                if x&y ==0:
                    ans = max(ans, bitmask_map[x] * bitmask_map[y])
        print(time.time()-t1)
        return ans