392. Is Subsequence leetcode binary search

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.

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?

方法一、

本题采用string的find_first_of的内置函数去实现，巧妙之处，使用了变量pos：起到了记录中间结果的作用，使得查找不必每次都从头开始。从而节约了时间

bool isSubsequence(string s, string t) {
int lenS = s.length();
int lenT = t.length();

if(lenT < lenS )
return false;

int pos = -1; //pos用来记录在t中查找s的每个字符的第一次字符出现的位置
int count = 0; //count表示每次遍历s的一个字符的位置下标
while(count < lenS)
{
char temp = s[count];
int loc = t.find_first_of(temp,pos+1); //下次得从pos+1的位置找temp第一个出现的位置，该函数如果查找失败则返回string::npos
if(loc == string::npos || loc < pos)
return false;
else
pos = loc ;//更新pos
count++;
}

return true;
}

方法二、思路：采用双指针，一次遍历即可，只要字符串t中含有s的所有元素即可。即在结束时判断指向s的指针是否指向末尾。

该解题思路有点类似349和350求两个数字数组的交集的思想，此处不再详述

bool isSubsequence(string s, string t) {
if(s.size()>t.size()) return false;
int len1 = s.size(), len2 = t.size(), k1 = 0, k2 = 0;
while(k1 < len1 && k2 < len2)
{
if(s[k1]==t[k2]) k1++;
k2++;
}
return k1 == len1;
}

方法三、动态规划等的思想，后续实现