392. 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.

一开始我的思路是这样子的：在子序列s中每一个字符都在序列t里面去找，在实现的过程中怎么都找不到错误的地方在哪里。附上代码：

class Solution {public:    bool isSubsequence(string s, string t) {        int lengths = s.length();        int lengtht = t.length();        int i = 0;        int j = 0;        bool flag = false;        if(lengths == 0)            return true;        while(lengths)        {            while(lengtht)            {                if(s[i] == t[j])                {                    flag = true;                    break;                }                else                {                    flag = false;                    j++;                }                lengtht--;            }            if(flag==false)                break;            lengths--;            i++;        }        return flag;    }};

然后发现有一点逻辑错误，经修改后变成：

class Solution {public:    bool isSubsequence(string s, string t) {        int lengths = s.length();        int lengtht = t.length();        int i = 0;        int j = 0;        bool flag = false;        if(lengths == 0)            return true;        while(lengths)        {            flag = false;            j = 0;            while(lengtht)            {                if(s[i] == t[j])                {                    flag = true;                    break;                }                j++;                lengtht--;            }            if(flag==false)                break;            lengths--;            i++;        }        return flag;    }};

发觉我的代码很啰嗦，于是简洁了一下：

class Solution {public:    bool isSubsequence(string s, string t) {        int lengths = s.length();        int lengtht = t.length();        int i = 0;        int j = 0;        bool flag = false;        if(lengths == 0)            return true;        for(;i<lengths;i++)        {            flag = false;            for(j = 0;j<lengtht;j++)            {                if(s[i] == t[j])                {                    flag = true;                    break;                }            }            if(flag==false)                break;        }        return flag;    }};

然后这个方法就卡在这个结果上：

Input:"acb""ahbgdc"

Output:true

Expected:false

我明白，原来我忽略了acb这个顺序，OH，MYGOD！！！！

最后我用了Python的all().可参考：all()

AC：

class Solution(object):    def isSubsequence(self, s, t):        """        :type s: str        :type t: str        :rtype: bool        """        t = iter(t)        return all(c in t for c in s)