力扣题:公共前缀/单词-11.18

力扣题-11.18

[力扣刷题攻略] Re：从零开始的力扣刷题生活

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;
    }
};