内容:
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).
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:
public class Solution {
public boolean isSubsequence(String s, String t) {
int sindex = 0, tindex = 0;
while(sindex < s.length() && tindex < t.length()) {
if(s.charAt(sindex) == t.charAt(tindex)) {
sindex++;
}
tindex++;
}
if(sindex == s.length()) return true;
return false;
}
}
本文介绍了一种简单有效的方法来判断一个字符串是否为另一个字符串的子序列。通过双指针技术,该方法能在较短的时间内得出结论,适用于s较短(<=100),t可能很长(~500,000)的情况。
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