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[LeetCode] 392. Is Subsequence Java
题目:
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
.
题意及分析:给出一个字符串s,给出一个字符串t,求s是否是t中的一个序列。就是要判断s中的每一个字符在t中是否顺序存在。这道题是没看答案作出的第一到贪心算法的题。。。以上面的Example作分析,在t中先找到字符a,那么在t中找到b,c的概率就更大;对于剩下的"bc",先在t中找到b,剩下的字符就越多,找到c的概率就越大。所以,这里我们对于s中的每一个字符都尽量在t中先找到,那么剩下的字符就越多,符合要求的概率就越大。
代码:
public class Solution { public boolean isSubsequence(String s, String t) { int sLength=s.length(); //s的长度 int tLength=t.length(); //t的长度 int sIndex=0; //记录找到了s的第几个字符 int tIndex=0; //记录遍历到了t的第几个字符 while(tIndex<tLength){ //这里用到的贪心算法,就是在t中越先找到 s的字符 那么在t中就更容易的找到 s剩下的字符 if(sIndex<sLength&&(s.charAt(sIndex)==t.charAt(tIndex))){ //找到一个相等的字符就查找s的下一个字符 sIndex++; } tIndex++; } if(sIndex==sLength){ //从前往后在t中找到了s的所有字符 。所以存在 return true; } else{ return false; } } }
[LeetCode] 392. Is Subsequence Java