LeetCode:Is Subsequence
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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
.
给出字符串s和字符串t,检查字符串s是否为t的子序列,假设两个字符串中均为小写字母。所谓子序列,即在原字符串中,去除某些位的字符,剩下的字符保持相对顺序不变,则形成原字符串的字序列。
解题思路1:
利用两个指针,一个指针1指向字符串s的首字符,一个指针2指向t的首字符,接下来进行遍历,若当前位置两个指针所指字符相等,则同时移动两个指针指向下一个字符,否则,只移动指向原字符串的指针,同时注意判断指针1是否到达字符串s的末尾,若是,则范围true。
代码如下:
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; }}
解题思路2:
利用Java的String提供的indexOf(char c, int index)方法,获取从index位置开始,字符c在字符串中的位置,若找不到则返回-1;经过测试,性能比使用双指针的方法更好。
代码如下:
public class Solution { public boolean isSubsequence(String s, String t) { if(t.length() < s.length()) return false; int prev = 0; for(int i = 0; i < s.length();i++) { char tempChar = s.charAt(i); prev = t.indexOf(tempChar,prev); if(prev == -1) return false; prev++; } return true; }}
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