代码随想录day8 ● 344.反转字符串 ● 541. 反转字符串II ● 剑指Offer 05.替换空格 ● 151.翻转字符串里的单词 ● 剑指Offer58-II.左旋转字符串

344. 反转字符串

class Solution {
public:
    void reverseString(vector<char>& s) {
        for (int i = 0, j = s.size() - 1; i < s.size() / 2; i++, j--){
            //交换两头的字母，经典倒油瓶
            //char temp = s[i];
            //s[i] = s[j];
            //s[j] = temp;

            //函数法
            swap(s[i], s[j]);
        }
    }
};

541. 反转字符串 II

class Solution {
public:

    void reverse(string& s, int start, int end){
        for (int i = start, j = end; i < j; i++, j--){
            swap(s[i], s[j]);
        }
    }

    string reverseStr(string s, int k) {
        for (int i = 0; i < s.size(); i += 2 * k){
            //反转前K个字母
            if (i + k <= s.size()){
                reverse(s, i, i + k - 1);
            //反转剩下的字母
            } else {
                reverse(s, i, s.size() -1);
            }
        }
        return s;
    }
};

剑指 Offer 05. 替换空格

class Solution {
public:
    string replaceSpace(string s) {
        //先统计空格的个数
        int count = 0;//空格个数
        int sOldSize = s.size();
        for (int i = 0; i < s.size(); i++){
            if (s[i] == ' '){
                count++;
            }
        }

        //扩充数组的大小
        s.resize(sOldSize + count * 2);
        int sNewSize = s.size();
        //从后往前将空格替换为%20
        for (int i = sOldSize - 1, j = sNewSize - 1; i < j; i--, j--){
            if (s[i] != ' '){
                s[j] = s[i];
            } else {
                s[j] = '0';
                s[j - 1] = '2';
                s[j - 2] = '%';
                j -= 2;
            }
        }
        return s;
    }
};

151. 反转字符串中的单词

class Solution {
public:
    //反转字符串
    void reverse(string& s, int start, int end){
        for (int i = start, j = end; i < j; i++, j--){
            swap(s[i], s[j]);
        }
    }
    //删除多余空格
    void  removeExtraSpaces(string& s){
        //定义慢指针
        int slow = 0;
        for (int i = 0; i < s.size(); ++i){
            //遇到非空格就处理，即删除所有的空格
            if (s[i] != ' '){
                if (slow != 0){
                    s[slow++] = ' ';
                }
                //补上该单词
                while (i < s.size() && s[i] != ' '){
                    s[slow++] = s[i++];
                }
            }
        }
        //重新设置数组大小
        s.resize(slow);
    }
    //反转数组内单词
    string reverseWords(string s) {
        //删除数组内多余空格
        removeExtraSpaces(s);
        //反转整个字符串
        reverse(s, 0, s.size() - 1);
        //第一个单词的下标为0
        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;
    }
};

剑指 Offer 58 - II. 左旋转字符串

class Solution {
public:
    string reverseLeftWords(string s, int n) {
        reverse(s.begin(), s.begin() + n);
        reverse(s.begin() + n, s.end());
        reverse(s.begin(), s.end());
        return s;
    }
};

class Solution {
public:
    string reverseLeftWords(string s, int n) {
        //记录要移动的字母
        string str = {s.begin(), s.begin() + n};
        //将后面的字母向前移动
        for (int i = n; i < s.size(); i++){
            s[i - n] = s[i];
        } 
        int j = 0;
        int k = s.size() - n;
        //将记录下来的字母替换原来的字母
        for (; k < s.size(); k++){
            s[k] = str[j];
            cout << str[j] << endl;
            j++;
        }
        return s;
    }
};