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?
【题目分析】
判断一个字符串是否是另一个字符串的子序列，如果s是t的子序列，t可以这样构成：在s字符串中的任意位置插入任意长度的字符串。
【思路】
1.维持两个字符串指针，分别指向s和t，如果当前字符相同，则指针都向后移动，否则只移动t的指针，直到s中出现的字符都在t中出现过了，我们可以判定s是t的子序列。代码如下：
1 public class Solution { 2 public boolean isSubsequence(String s, String t) { 3 int sindex = 0, tindex = 0; 4 while(sindex < s.length() && tindex < t.length()) { 5 if(s.charAt(sindex) == t.charAt(tindex)) { 6 sindex++; 7 } 8 tindex++; 9 } 10 if(sindex == s.length()) return true; 11 return false; 12 } 13 }
2.我们对上面的代码进行改进，代码如下：
1 public class Solution { 2 public boolean isSubsequence(String s, String t) { 3 if(t.length() < s.length()) return false; 4 int prev = 0; 5 for(int i = 0; i < s.length();i++) { 6 char tempChar = s.charAt(i); 7 prev = t.indexOf(tempChar,prev); 8 if(prev == -1) return false; 9 prev++; 10 } 11 12 return true; 13 } 14 }
可以看到这两种方法的差别很大。之所以有这样的差别，是因为在第一种方法中我们每查看一个字符就要调用一次charAt()方法。而在第二种方法中使用indexOf()方法可以直接跳过不匹配的字符，这样大大减少的了函数的调用次数，减少时间复杂度，简直太棒了！