标签:origin example osi without length not case tco code
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;
}
};
标签:origin example osi without length not case tco code
原文地址:https://www.cnblogs.com/A-Little-Nut/p/10061315.html