392. Is Subsequence

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本文介绍了一个用于检查一个字符串是否为另一个字符串子序列的有效算法。通过迭代比对的方式，该算法能够在较短的时间内完成判断，适用于处理大量字符串的场景。

题目：392. Is Subsequence

Total Accepted: 28646
Total Submissions: 64661
Difficulty: Medium
Contributor: LeetCode

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.

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?

解题思路：

s1，s2分别为s和t的长度，s的第一位在t中的位置的范围为0~s2-s1，s的第二位在t中的位置的范围为第一位的位置+1~s2-s1+1。s中的每一位的位置范围都可以按照这个规则递推下去。对于s中的每一位，在t中这个范围中搜索，若找到则继续搜索下一位，若没找到则直接返回false。直到每一位都找到了则返回true。

代码：

class Solution {
public:
    bool isSubsequence(string s, string t) {
    	bool check = false;
        int s1 = s.size();
        int s2 = t.size();
        if(s1 == 0)
        	return true;
        else if(s2 == 0)
        	return false;
        int begin = 0;
        int end = s2 - s1;
        for(int i = 0; i < s1; i++){
        	check = false;
        	for(int j = begin; j < end + 1; j++){
        		if(s[i] == t[j]){
        			check = true;
        			begin = j + 1;
        			end ++;
        			break;
        		}
        	}
        	if(!check)
        		return false;
        }
        return check;
    }
};

做完之后看了solution里面的代码发现同样的思路能够用更简洁的代码来实现，我的代码有一点复杂了。参考代码（java）如下：

public class Solution {
    public boolean isSubsequence(String s, String t) {
        if (s.length() == 0) return true;
        int indexS = 0, indexT = 0;
        while (indexT < t.length()) {
            if (t.charAt(indexT) == s.charAt(indexS)) {
                indexS++;
                if (indexS == s.length()) return true;
            }
            indexT++;
        }
        return false;
    }
}