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.
分析
这道题我先是用动态规划和二维数组来做的，但空间不够
class Solution {
public:
bool isSubsequence(string s, string t) {
int dp[s.size()+1][t.size()+1];
for(int i=0;i<=t.size();i++)
dp[0][i]=1;
for(int i=1;i<=s.size();i++){
dp[i][0]=0;
for(int j=1;j<=t.size();j++)
if(s[i-1]==t[j-1])
dp[i][j]=dp[i-1][j-1]==1?1:0;
else
dp[i][j]=dp[i][j-1];
}
if(dp[s.size()][t.size()]==1)
return true;
else
return false;
}
};
接着，我压缩了空间，变为一维数组，虽然ac了，并且可以继续在此基础继续做一些优化，但效果不明显。
class Solution {
public:
bool isSubsequence(string s, string t) {
int dp[t.size()+1];
for(int i=0;i<=t.size();i++)
dp[i]=1;
for(int i=1;i<=s.size();i++){
int t1=dp[0];
dp[0]=0;
for(int j=1;j<=t.size();j++){
int t2=dp[j];
if(s[i-1]==t[j-1])
dp[j]=t1==1?1:0;
else
dp[j]=dp[j-1];
t1=t2;
}
}
if(dp[t.size()]==1)
return true;
else
return false;
}
};
本着不断优化的想法，还可以不用动态规划，其实这也是最容易想到的办法，但因为最近在集中训练动态规划，所以把这个办法放到了最后
class Solution {
public:
bool isSubsequence(string s, string t) {
int index=0;
for(int i=0;i<t.size()&&index<s.size();i++)
if(s[index]==t[i])
index++;
if(index==s.size())
return true;
else
return false;
}
};