LeetCode刷题（19）

Substring with Concatenation of All Words
寻找包含所有单词的子字符串，按照人工寻找的方式有如下代码，对于每一个index依次进行比对，该方案通过了167/169，最后两个字符串超时了

class Solution(object):    def findSubstring(self, s, words):        """        :type s: str        :type words: List[str]        :rtype: List[int]        """        ls = len(s)        lw = len(words)        if not ls or not lw:            return -1        wl = len(words[0])        res = []        dic = {}        for word in words:            if not dic.has_key(word):                dic[word] = 1            else :                dic[word] += 1        i = 0         while i + wl <= ls:            if s[i: i+wl] not in words:                i += 1            else :                tdic = {}                tdic[s[i: i+wl]] = 1                j = 1                while i + j * wl + wl <= ls:                    if s[i + j * wl: i + j * wl + wl] not in words:                        #i = i + 1                        break                    else :                        if not tdic.has_key(s[i + j * wl: i + j * wl + wl]):                            tdic [s[i + j * wl: i + j * wl + wl]] = 1                            j += 1                        else :                            if tdic[s[i + j * wl: i + j * wl + wl]] < dic[s[i + j * wl: i + j * wl + wl]]:                                tdic [s[i + j * wl: i + j * wl + wl]] += 1                                j += 1                            else :                                break                flag = 0                for key in dic:                    if not tdic.has_key(key):                        flag = 1                        break                    elif tdic[key] != dic[key]:                        flag = 1                        break                if not flag :                    res += [i]                i += 1        return res

研究一下讨论区的其他算法：

class Solution(object):    def findSubstring(self, s, words):        """        :type s: str        :type words: List[str]        :rtype: List[int]        """        if not s or not words or not words[0]:            return []        n = len(s)        k = len(words[0])        t = len(words) * k        req = {}        for w in words:            req[w] = req[w] + 1 if w in req else 1        ans = []        for i in range(min(k, n - t + 1)):            self._findSubstring(i, i, n, k, t, s, req, ans)        return ans    def _findSubstring(self, l, r, n, k, t, s, req, ans):        curr = {}        while r + k <= n:            w = s[r:r + k]            r += k            if w not in req:                l = r                curr.clear()            else:                curr[w] = curr[w] + 1 if w in curr else 1                while curr[w] > req[w]:                    curr[s[l:l + k]] -= 1                    l += k                if r - l == t:                    ans.append(l)

这个算法实现的主要是一个滑动窗口，当word[0],word[1],word[0]出现时，向后滑一个从word[1],word[0]以及往后开始重新记，其中假设word[0]在words中是唯一的，不唯一时就在超出数量时滑动。为了防止每次滑动一个单词长度遗漏答案，特意将开始的word长度字符分别做头。

AC算法看起来正是用来解决此类问题的，AC算法详解