Leetcode String 知识点总结

• 551. Student Attendance Record I：一堂课不合格的条件是：两次及以上的缺席（‘A‘）或连续三次的迟到（‘L‘），给定一个字符串，求该学生是否合格，Easy

class Solution(object):
    def checkRecord(self, s):
        """
        :type s: str
        :rtype: bool
        """
        countA = 0
        for i in range(len(s)):
            if s[i]=='A':
                countA+=1
                if countA>1:
                    return False
            if i>1 and s[i]=='L' and s[i-1] == 'L' and s[i-2]=='L':
                return False
        return True

• 552. Student Attendance Record II：求n长满足学生出勤合格的字符串的总个数，Hard

设dp[k][x][y] 为长度为k的有x个A和以y个L结尾的个数
dp[n][0][0] = sum(dp[n-1][0]) 即结尾增加一个P
dp[n][0][1] = dp[n-1][0][0] 即结尾增加一个L
dp[n][0][2] = dp[n-1][0][1] 即结尾增加一个L
dp[n][1][0] = sum(dp[n-1][0]) + sum(dp[n-1][1]）前者增加一个A后者增加一个P
dp[n][1][1] = dp[n-1][1][0] 即结尾增加一个L
dp[n][1][2] = dp[n-1][1][1] 即结尾增加一个L

由于dp[n] 只和dp[n-1]有关，所以可以采用滚动数组，该思路参考书影博客

class Solution(object):
    def checkRecord(self, n):
        """
        :type n: int
        :rtype: int
        """
        MOD = 1000000007
        dp = [[1,1,0],[1,0,0]]
        for i in range(2,n+1):
            tmp = sum(dp[0]) % MOD
            dp[0][2] = dp[0][1]
            dp[0][1] = dp[0][0]
            dp[0][0] = tmp

            tmp = (dp[0][0] + sum(dp[1]))%MOD
            dp[1][2] = dp[1][1]
            dp[1][1] = dp[1][0]
            dp[1][0] = tmp

        return (sum(dp[0])+sum(dp[1]))%MOD

• 504. Base 7：求一个整数的7进制表达，Easy
  

注意正负号和0的特殊处理

class Solution(object):
    def convertToBase7(self, num):
        """
        :type num: int
        :rtype: str
        """
        ret,flag = '',1
        if num<0:
            flag = 0
            num = -num

        while(num):
            ret = str(num % 7)+ret
            num = num/7
        if ret == '':
            return '0'
        if flag: return ret
        else:
            return '-'+ret

• 809. Expressive Words：判断words列表中有多少个word能够扩展成S, Medium
  

能够扩展的条件是 单词连续个数大于3
先统计S中单词的个数 query 如heeellooo -> [(h,1),(e,3),(l,2),(o,3)]
对于每一个word来说，也统计word中单词的个数，candidate 如 hello -> [(h,1),(e,1),(l,2),(o,1)]
检查 len(query) == len(candidate) ?
检查 query[i][0] == candidate[i][0] ?
检查 query[i][1] >= candidate[i][1] 且如果> 成立的话 ，需要检查query[i][1]>=3 ?

class Solution(object):
    def expressiveWords(self, S, words):
        """
        :type S: str
        :type words: List[str]
        :rtype: int
        """
        def countLetters(word):
            cnt,last,ans = 0,'',[]
            for c in word:
                if c == last:
                    cnt+=1
                else:
                    if cnt: ans.append((last,cnt))
                    cnt,last = 1,c
            if cnt: ans.append((last,cnt))
            return ans

        def check(query,candidate):
            if len(query)!=len(candidate): return 0
            for c,w in zip(query,candidate):
                if  c[0] != w[0]: return 0
                if  c[1] < w[1]: return 0
                if  c[1] > w[1] and c[1]<3: return 0
            return 1

        query = countLetters(S)
        ans = 0
        for word in words:
            candidate = countLetters(word)
            ans+=check(query,candidate)
        return ans