- Number of Matching Subsequences Medium
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Share Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.
Example : Input: S = "abacde" words = ["a", "bb", "acd", "ace"] Output: 3 Explanation: There are three words in words that are a subsequence of S: "a", "acd", "ace". Note:
All words in words and S will only consists of lowercase letters. The length of S will be in the range of [1, 50000]. The length of words will be in the range of [1, 5000]. The length of words[i] will be in the range of [1, 50].
思路：用字典保存S字符串每个字符出现下标，例子：dic['a']=[0,2],dic['b']=[1] 然后遍历words，置标示位pre=-1，遍历每个字符c，到相应的dic[c]中查找最近的位置，各种判断。。。
代码：python3
class Solution:
def numMatchingSubseq(self, S: str, words) -> int:
dic = {}
for (index,value) in enumerate(S):
if value in dic.keys():
dic[value].append(index)
else:
dic[value] = [index]
print(words)
delList = []
for word in words:
pre = -1
print("start "+word)
needJump = False
for (index,value) in enumerate(word):
if needJump == True:
needJump = False
break
if value not in dic.keys():
delList.append(word)
needJump = True
continue
b = False
for index in dic[value]:
if index > pre:
b = True
pre = index
break
if b == False:
if word in words:
delList.append(word)
needJump = True
break
print(delList)
return len(words)-len(delList)
if __name__ == '__main__':
print(Solution().numMatchingSubseq("dcoegauiqk",["dco","coegauiq","ehqyopnqgm","wtqbgozlae"]))
