392. Is Subsequence
2018-03-30 19:23
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判断S是否是T的一部分:
1) 若s.size()>t.size(),则return false;
2) 否则,从s和t首部开始遍历,若对应位置的字符相等,则s,t下标同时后移一位;否则,t下标后移一位class Solution {
public:
bool isSubsequence(string s, string t) {
if(s.size()> t.size())
return 0;
int flag=0;
for(int i=0,j=0;i<s.size()&&j<t.size();)
{
if(s[i]==t[j])
{
i++;
j++;
flag++;
}
else
{
j++;
}
}
if(flag==s.size())
return 1;
else
return 0;
}
};
1) 若s.size()>t.size(),则return false;
2) 否则,从s和t首部开始遍历,若对应位置的字符相等,则s,t下标同时后移一位;否则,t下标后移一位class Solution {
public:
bool isSubsequence(string s, string t) {
if(s.size()> t.size())
return 0;
int flag=0;
for(int i=0,j=0;i<s.size()&&j<t.size();)
{
if(s[i]==t[j])
{
i++;
j++;
flag++;
}
else
{
j++;
}
}
if(flag==s.size())
return 1;
else
return 0;
}
};
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