leetcode392 Is Subsequence

"""
 Given a string s and a string t, check if s is subsequence of t.
You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).
Example 1:
s = "abc", t = "ahbgdc"
Return true.
Example 2:
s = "axc", t = "ahbgdc"
Return false.
"""
"""
自强自信题。
看到这道题的想法就是 在子串中遍历每个字符，
按顺序判断是否在父串中
如果在，flag为True，父串变为所匹配字母的下标+1到末尾
如果不在，flag为False，跳出循环
"""
class Solution1(object):
    def isSubsequence(self, s, t):
        n = len(s)       #子串长度
        if n == 0:
            return True
        flag = False
        for i in range(n):
            if s[i] in t:                #判断字符是否在t里
                t = t[t.index(s[i])+1:]  #bug没有加1
                flag = True
            else:
                flag = False
                break            #bug没有提前跳出，后面的字符如果匹配成功，最后会是True
        return flag

"""
解法二：提供一种队列法。与Solution1思想一致。
这里用队列十分高级
"""
class Solution2(object):
    def isSubsequence(self, s, t):
        from collections import deque
        queue = deque(s)          #deque模块提供了两端都可以操作的序列，
                                  # 这意味着，在序列的前后你都可以执行添加或删除操作
        for c in t:
            if not queue:
                return True
            if c == queue[0]:       # d = deque('12345')
                queue.popleft()     # d.popleft()  '1'   extendleft
        return not queue            # d.pop()      '5'   extend

"""
第三种是种双指针法，很重要，许多题目都可以使用
双指针要用while循环，单独操作i, j步长，总长度为终止条件
"""
class Solution3(object):
    def isSubsequence(self, s, t):
        slen = len(s)
        tlen = len(t)
        i, j = 0, 0                      #bug代码
        while(i < slen and j < tlen):    #for i in range(slen):
            if(s[i] == t[j]):            #    for j in range(tlen):
                i += 1                   #        if s[i] == t[i]:
            j += 1
        return i == slen