代码随想录第八天|151.翻转字符串里的单词、卡码网：55.右旋转字符串、28. 实现 strStr()、459.重复的子字符串-优快云博客

151.翻转字符串里的单词

思路：先去空格，翻转整个字符串，然后翻转每个单词。
题解：

class Solution {
public:
    void reverse(string &s, int left, int right)
    {
        while (left < right)
        {
            swap(s[left], s[right]);
            ++left;
            --right;
        }
    }
    string reverseWords(string s) {
        int slow = 0;
        for (int i = 0; i < s.size(); ++i)
        {
            if (s[i] != ' ')
            {   
                if (slow != 0) s[slow++] = ' ';
                while(s[i] != ' ' && i < s.size()) s[slow++] = s[i++];
            }
        }
        s.resize(slow);
        reverse(s, 0, s.size() - 1);
        int start = 0;
        for (int i = 0; i <= s.size(); i++)
        {
            if (i == s.size() || s[i] == ' ')
            {
                reverse(s, start, i-1);
                start = i + 1;
            }
        }
        return s;
    }
};

卡码网：55.右旋转字符串

题目链接：右旋转字符串
思路：先翻转整个字符串，然后分别翻转两段字符。
题解：

# include<iostream>
# include<string>
using namespace std;
void reverse(string &s, int left, int right)
{
    while(left < right)
    {
        swap(s[left], s[right]);
        ++left;
        --right;
    }
}
int main(){
    int k;
    string s;
    cin >> k;
    cin >> s;
    reverse(s, 0, s.size()-1);
    reverse(s, 0, k-1);
    reverse(s, k, s.size()-1);
    cout << s;
}

28. 实现 strStr()

思路：考察KMP，需要知道如何生成前缀表。
题解：

class Solution {
public:
    void getNext(const string &s, int *next)
    {
        int j = 0;
        next[0] = j;
        for (int i = 1; i < s.size(); ++i)
        {
            while(j > 0 && s[j] != s[i]) j = next[j-1];
            if (s[j] == s[i]) ++j;
            next[i] = j;
        }
    }
    int strStr(string haystack, string needle) {
        vector<int> next(needle.size());
        getNext(needle, &next[0]);
        int j = 0;
        for (int i = 0; i < haystack.size(); ++i)
        {
            while(j > 0 && haystack[i] != needle[j])    j = next[j-1];
            if (haystack[i] == needle[j])   ++j;
            if (j == needle.size()) return  i + 1 - needle.size();
        }
        return -1;
    }
};

459.重复的子字符串

思路：同上
题解：

class Solution {
public:
   void getNext(const string &s, int *next)
   {
       int j = 0;
       next[0] = j;
       for (int i = 1; i < s.size(); ++i)
       {
           while(j > 0 && s[j] != s[i]) j = next[j-1];
           if (s[j] == s[i])   ++j;
           next[i] = j;
       }
   }
   bool repeatedSubstringPattern(string s) {
       vector<int> next(s.size());
       getNext(s, &next[0]);
       if (next[s.size() - 1] != 0)
       {
           int sublen = s.size() - next[s.size() - 1];
           if (s.size() % sublen == 0)
               return true;
       }
       return false;
   }
};