Leetcode

1. IsSubsequence
   Python
   
class Solution(object):
def isSubsequence(self, s, t):
"""
:type s: str
:type t: str
:rtype: bool
"""
for i, ele in enumerate(s):
if ele in t:
pos = t.index(ele)
t = t[pos+1:]
else:
return False
return True

   C
   
bool isSubsequence(char* s, char* t) {
int ls, lt,i,j,start=0;
ls = strlen(s);
lt = strlen(t);
for (i = 0; i < ls; i++) {
if (start >= lt)
return false;
for (j = start; j < lt; j++) {
if (s[i] == t[j]) {
start = j + 1;
break;
}
else if (j >= lt - 1)
return false;
}
}
return true;
}

   大神就是大神

bool isSubsequence(char* s, char* t) {    while (*t)        s += *s == *t++;    return !*s;}

如果s与t匹配，则*s==*t++为1，所以s的指针加1。主要思路就是遍历t与s中字符匹配，匹配到则指针加1，当t遍历完，检查s是否还剩有元素。对c的指针用法还是不熟练