392. 判断子序列

一、题目描述

二、示例

三、难度

简单

四、代码

Java版

class Solution {
    public boolean isSubsequence(String s, String t) {
        int from = 0;
        for (int i = 0; i < s.length(); ++i) {
            from = t.indexOf(s.charAt(i), from);
            if (from == -1) return false;
            ++from;
        }
        return true;
    }
}

Python版

class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        from_index = -1
        for ch in s:
            if t.find(ch, from_index + 1) != -1:
                from_index = t.find(ch, from_index + 1)
            else:
                return False
        return True