代码随想录算法训练营第9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

本日任务很重，加油。

151.翻转字符串里的单词

class Solution {
public:
    void reverse(string& s, int start, int end){
        while(end > start){
            swap(s[end], s[start]);
            end--;
            start++;
        }
    }
    string reverseWords(string s){ 
        int n = s.size();
        int left = 0, right = n - 1;
        // 去除开始的空格
        while(left <n && s[left] == ' '){
            left++;
        }
        // 去除最后的空格
        while(right >= 0 && s[right] == ' '){
            right--;
        }
        int new_index = 0;
        for(int i = left; i <= right; i++){
            if(i > 0 && s[i] == s[i - 1] && s[i] == ' '){
                continue;
            }else{
                s[new_index++] = s[i];
            }
        }
        s.resize(new_index);
        n = s.size();
        reverse(s, 0, n - 1);
        for(int start = 0, end = 0; end < n; end++){
            if(s[end] == ' '){
                reverse(s, start, end - 1);
                start = end + 1;
            }
            if(end == n - 1){
                reverse(s, start, end);
            }
        }
        return s;
    }
};

55.右旋转字符串

55. 右旋字符串（第八期模拟笔试） (kamacoder.com)

#include<iostream>
#include<algorithm>

using namespace std;
void reverse_str(string& s, int start, int end){
    while(start < end){
        swap(s[start], s[end]);
        start++;
        end--;
    }
} 

int main(){
    string s;
    int k;
    cin >> s;
    cin >> k;
    int n = s.size();
    reverse_str(s, 0, n - 1);
    reverse_str(s, 0, k - 1);
    reverse_str(s, k, n - 1);
    cout << s << endl;
    return 0;
}

28. 实现 strStr()

28. 找出字符串中第一个匹配项的下标 - 力扣（LeetCode）

kmp 懂得都懂，太难了。

class Solution {
public:
    vector<int> getNext(string needle){
        int n = needle.size(), prehead = 0;
        vector<int> next;
        next.push_back(0);
        for(int i = 1; i < n; i++){
            while(needle[i] != needle[prehead]){
                if(prehead == 0){
                    break;
                }else{
                    prehead = next[prehead - 1];
                }
            }
            if(needle[i] == needle[prehead]) prehead++;
            next.push_back(prehead);
        }
        return next;
    }
    int strStr(string haystack, string needle) {
        int n = haystack.size(),  m = needle.size();
        vector<int> next = getNext(needle);
        for(int i = 0, j = 0; i < n; i++){
            while(j > 0 && haystack[i] != needle[j]) j = next[j - 1];
            if(haystack[i] == needle[j]) j++;
            if(j == m) return i - j + 1;
        }
        return -1;
    }
};

459.重复的子字符串

459. 重复的子字符串 - 力扣（LeetCode）

class Solution {
public:
    vector<int> getNext(string s){
        vector<int> next;
        next.push_back(0);
        int j = 0, n = s.size();
        for(int i = 1; i < n; i++){
            while(j > 0 && s[i] != s[j]) j = next[j - 1];
            if(s[i] == s[j]) j++;
            next.push_back(j);
        }
        return next;
    }
    bool repeatedSubstringPattern(string s) {
        vector<int> next = getNext(s);
        string t = s + s;
        int j = 0, n = t.size();
        // for(int i = 1; i < n - 1; i++){
        //     while(j > 0 && t[i] != s[j]) j = next[j - 1];
        //     if(t[i] == s[j]) j++;
        //     if(j == n / 2) return true;
        // }
        return next[n / 2 - 1] != 0 && n / 2 % (n / 2 - next[n / 2 - 1]) == 0;
    }
};