1. 字符串相加

1. 题目信息：
   
2. 题目连接：
   字符串相加

class Solution 
{
public:string addStrings(string num1, string num2) {int ent = 0;int count1 = 0;int count2 = 0;string sum;auto it1 = num1.rbegin();auto it2 = num2.rbegin();while(it1 < num1.rend() || it2 < num2.rend()){if(it1 < num1.rend()){count1 = *it1 -'0';}if(it2 < num2.rend()){count2 = *it2 - '0';}count1 = count1 + count2 + ent;ent = count1 / 10;count1 %= 10;sum.push_back(count1 + '0');count1 = 0;count2 = 0;if(it1 < num1.rend()){it1++;}if(it2 < num2.rend()){it2++;}}if(ent){sum.push_back(ent + '0');}auto it3 = sum.begin();auto it4 = sum.end() - 1;while(it3 < it4){swap(*it3,*it4);it3++;it4--;}return sum;}
};

2. 反转字母

1. 题目信息：
   
2. 题目连接：
   反转字母

class Solution 
{
public:string reverseOnlyLetters(string s) {//快排hoare法单趟string ret(s);int left = 0;int right = ret.size() - 1;while(left < right){while(left < right && !isalpha(ret[right])){right--;}while(left < right && !isalpha(ret[left])){left++;}swap(ret[left], ret[right]);left++;right--;}return ret;}
};

3. 字符串中唯一字母

1. 题目信息：
   
2. 题目连接：
   字符串中唯一字母
3. 思路：暴力求解

class Solution 
{
public:int firstUniqChar(string s) {for (int i = 0; i < s.size(); i++){int flag = 1;//暴力查找for(int k = 0; k < s.size();){if(k == i){k++;continue;}//漏掉前面的字符if(s[i] == s[k]){flag = 0;break;}k++;}//找到存在唯一字符if (flag){return i;}}return -1;}
};

4. 字符串中最后一个单词

1. 题目信息：
   
2. 题目链接：
   字符串最后一个单词的长度

int main() 
{string str;//遇到' '字符不停读取，一次读取一行getline(cin, str);int cur = str.length() - 1;int i = 0;while(str[cur] != ' ' && cur >= 0){cur--;i++;}cout << i;return 0;
}

5. 验证回文串

1. 题目信息：
   2. 题目链接：
   验证回文串

class Solution 
{
public:bool isPalindrome(string s) {string copy_s(s);auto e = copy_s.begin(); while(e < copy_s.end()){if(*e >= 'A' && *e <= 'Z'){*e -= 'A' - 'a';}if((*e < 'a' || *e > 'z') && (*e < '0' || *e > '9')){//迭代器失效，漏掉一个位置copy_s.erase(e);}else{e++;}}//reverse//isalnum字符是否为大小写字母或者数字//tolower转小写 toupper转大写//islower isupperauto it1 = copy_s.begin();auto it2 = copy_s.rbegin();while(it1 < copy_s.end()){if(*it1 != *it2){return false;}it1++;it2++;}return true;}
};

6. 反转字符II

1. 题目信息：
   
2. 题目链接：
   反转字符II
3. 思路：一次向后遍历2k个字符，字符数大于k个逆置前k个，不足2k个整体逆置

class Solution 
{
public:string reverseStr(string s, int k) {string rs(s);int len = rs.length();for(int i = 0; i < len; i += 2 * k){reverse(rs.begin() + i, rs.begin() + min(i + k, len));}return rs;}
};

7. 反转字符串中的单词

1. 题目信息：
   
2. 题目链接：
   反转字符串中的单词

class Solution 
{
public:string reverseWords(string s) {string rs(s);int n = rs.length();int count = 0;for(int i = 0; i < n; i++){if(rs[i] == ' '){reverse(rs.begin() + count, rs.begin() + i);count = i + 1;}if(i == n - 1){reverse(rs.begin() + count, rs.end());}}return rs;}
};

8. 字符串相乘

1. 题目信息：
   
2. 题目链接：
   字符串相乘
3. 思路：模拟乘法步骤

class Solution 
{
public:string addStrings(string num1, string num2){int ent = 0;int count1 = 0;int count2 = 0;string sum;auto it1 = num1.rbegin();auto it2 = num2.rbegin();while (it1 < num1.rend() || it2 < num2.rend()){if (it1 < num1.rend()){count1 = *it1 - '0';}if (it2 < num2.rend()){count2 = *it2 - '0';}count1 = count1 + count2 + ent;ent = count1 / 10;count1 %= 10;sum.push_back(count1 + '0');count1 = 0;count2 = 0;if (it1 < num1.rend()){it1++;}if (it2 < num2.rend()){it2++;}}if (ent){sum.push_back(ent + '0');}auto it3 = sum.begin();auto it4 = sum.end() - 1;while (it3 < it4){swap(*it3, *it4);it3++;it4--;}return sum;}string multiply(string num1, string num2){//特殊处理if(num1 == "0" || num2 == "0"){return "0";}//模拟乘法的过程int size1 = num1.size();int size2 = num2.size();string ret;//num1 乘 num2for (int i = size1 - 1; i >= 0; i--){string tran;int ent = 0;int part1 = num1[i] - '0';int k = i;for (int j = size2 - 1; j >= 0; j--){//对齐while(k < size1 - 1){tran.push_back('0');k++;}int part2 = num2[j] - '0';int sum = part1 * part2 + ent;int bit = sum % 10;ent = sum / 10;tran.push_back(bit + '0');}//处理进位if (ent){tran.push_back(ent % 10 + '0');ent /= 10;}reverse(tran.begin(), tran.end());ret = addStrings(ret, tran);}return ret;}
};