力扣题-11.18
力扣题1：14.最长公共前缀
解题思想：先找到最小的字符串长度，然后进行字符串的遍历即可
class Solution(object):
def longestCommonPrefix(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
temp=''
minlen=len(strs[0])
for i in range(len(strs)):
if len(strs[i])<minlen:
minlen = len(strs[i])
for i in range(minlen):
now = strs[0][i]
flag=1
for j in range(len(strs)):
if strs[j][i]!=now:
flag = 0
break
if flag ==1:
temp =temp+now
elif flag ==0:
break
return temp
class Solution {
public:
string longestCommonPrefix(vector<string>& strs) {
string temp = "";
int minlen = strs[0].size();
for(int i=0;i<strs.size();i++){
if(strs[i].size()<minlen){
minlen = strs[i].size();
}
}
for(int i=0;i<minlen;i++){
char now = strs[0][i];
int flag = 1;
for(int j=0;j<strs.size();j++){
if(strs[j][i]!=now){
flag = 0;
break;
}
}
if(flag ==1){
temp =temp+now;
}
else{
break;
}
}
return temp;
}
};
力扣题2：434.字符串中的单词数
解题思想：进行遍历即可
class Solution(object):
def countSegments(self, s):
"""
:type s: str
:rtype: int
"""
temp = s.split()
return len(temp)
class Solution {
public:
int countSegments(string s) {
int result = 0;
int flag=0;
for(int i=0;i<s.size();i++){
if(s[i]==' ' && flag ==1){
flag=0;
}
else if(s[i]!=' ' && flag==0){
result++;
flag=1;
}
else if(s[i]!=' '&& flag==1){
continue;
}
else if(s[i]==' ' && flag==0){
continue;
}
}
return result;
}
};
