392[Medium]: Is Subsequence

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#C++ #字符串匹配

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本文介绍了一个经典的字符串匹配问题——检查一个字符串是否为另一个字符串的子序列，并提供了一种使用双循环进行暴力匹配的方法。该方法适用于短字符串s与长字符串t的匹配场景。

Part1:问题描述

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 sis 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?

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

Part2:解题思路

这一题我用的暴力解题，直接2个for循环判断s里的字符有没有在t中出现，值得注意的是字符出现的顺序也要一致。我觉得我有改进的地方在于第二个for循环的时候我用的下标是一个函数里面的全局变量index，这样就可以保证下一次从t中开始找的字符是上一次判断用过的字符之后的。第一个if里面有嵌套的一个if是因为不加类似第一个样例--最后一个相等的字符在最后出现时的样例判断会出错。

Part3:代码

class Solution {
public:
    bool isSubsequence(string s, string t) {
	int length1 = s.length();
	int length2 = t.length();
	if (length1 > length2) return false;
	int index = 0;
	int i = 0;
	for (i; i < length1; i++) {
		for (index; index < length2; ) {
			if (s[i] == t[index++]) {
				if (index == length2 && i != length1 - 1) {
					return false;
				}
				break;
			}
			
			if (index == length2) {
				return false;
			}
		}
	}
	return true;
}
};