392. Is Subsequence

题目：

Given a string s and a string t, check if s is subsequence of t.

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).

Follow up:
If there are lots of incoming S, say S1, S2, ... , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

Example 1:

Input: s = "abc", t = "ahbgdc"
Output: true

Example 2:

Input: s = "axc", t = "ahbgdc"
Output: false

Constraints:

• 0 <= s.length <= 100
• 0 <= t.length <= 10^4
• Both strings consist only of lowercase characters.

思路：

题意是给定两个字符串，s和t，要求确认s是不是t的子序列，即可以不连续。因此只要找到s是不是按序出现在t中即可。用一个变量当作s的index，然后变量t，如果有相同字母，s的index前进一位，最后只要看index是不是等于s的长度即可，如果相等则说明s完整的出现了，否则没有完整的出现。这里注意如例子中，如果s已经走完, index=3时，此时的s[3]是可以访问的，char应该为''，因为t中没有这个char，所以index会保持3不动直到循环结束。但是如果s[4]是会报错的，因为数组越界。

代码：

class Solution {
public:
    bool isSubsequence(string &s, string &t) {
        int index=0;
        for(int i =0;i<t.size();i++)
        {
            if(s[index]==t[i])
                index++;
        }
        return index==s.size();
    }
};

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