代码随想录day09

151.反转字符串中的单词，需二刷

//先去除多余空格，再反转所有字符，再反转单词，即可反转字符串中的单词

    void removeWhiteSpace(string& s){
        int slowIndex = 0;
        for(int fastIndex = 0; fastIndex < s.size(); fastIndex++){
            if(s[fastIndex] != ' '){
                if(slowIndex != 0){
                    s[slowIndex++] = ' ';
                }
                while(fastIndex < s.size() && s[fastIndex] != ' '){
                    s[slowIndex++] = s[fastIndex++];
                }
            }
        }
        s.resize(slowIndex);
    }
    void swapword(string& s, int left, int right){
        for(; left < right; left++, right--){
            swap(s[left], s[right]);
        }
    }
    string reverseWords(string s) {
        removeWhiteSpace(s);
        swapword(s, 0, s.size()-1);
        int start = 0;
        for(int i = 0; i <= s.size(); i++){
            if(s[i] == ' ' || i == s.size()){
                swapword(s,start,i-1);
                start = i + 1;
            }
        }
        return s;
    }

55.右旋字符串

//了解反转，再反转的用法

#include<iostream>
#include<algorithm>
using namespace std;

int main(){
    int k;
    string s;
    cin >> k;
    cin >> s;
    reverse(s.begin(),s.end());
    reverse(s.begin(), s.begin()+k-1);
    reverse(s.begin()+k,s.end());
    cout << s;
}

28.找出字符串中第一个匹配项的下标，需二刷

//KMP经典问题，需了解KMP算法中的具体步骤及实现细节，主要了解最长公共前后缀的概念，以及字符不匹配时从j以前字符串的最长公共前后缀位置开始重新匹配

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

459.重复的子字符串，需二刷

//解法1 KMP算法，需了解为何最长公共前后缀不包含的字符串什么时候是代表s的最小重复子串

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

//解法2.移动匹配

//如果由重复子串构成，则能在拼接出的字符串中找到新的相同字符串

    bool repeatedSubstringPattern(string s) {
        string t = s + s;
        t.erase(t.begin()); 
        t.erase(t.end() - 1); // 掐头去尾
        if (t.find(s) != std::string::npos) return true; // r
        return false;
    }