1.题目截图（把字符串转换成整数----atoi）

1.1题目解析（在代码里）

class Solution {
public:
    int myAtoi(string str) {
        // 100%  97.45%   
        int len = str.size();
        if(len == 0)
            return 0;
        int i = 0, flag = 1, isSignal = 0, res = 0;
        while(i < len && str[i] == ' ')  //跳过空格
            i++;
        if(i >= len)  //如果全是空格
            return 0;
        while(i < len && str[i] == '-' || str[i] == '+'){  //取符号
            if(str[i] == '-')
                flag = -1;
            i++;
            isSignal++;
            if(isSignal > 1)  //不能超过一个符号
                return 0;
        }
        while(isdigit(str[i])){  //只要前面的那些数字
            if(res > INT_MAX / 10 || (res == INT_MAX / 10 && str[i] > '7'))  //用int来控制不溢出
                return flag == 1? INT_MAX : INT_MIN;
            res = res * 10  - '0' + str[i];
            i++;
        } 
        return flag * res;
    }
};

2.Leetcode415--字符串相加

class Solution {
public:
    string addStrings(string num1, string num2)
    {
        int end1 = num1.size() - 1;
        int end2 = num2.size() - 1;
        string str;
        int next = 0;     //用来进位
        while(end1 >= 0 || end2 >= 0)
        {
            int x1 = end1 >= 0 ? num1[end1--] - '0' : 0;
            int x2 = end2 >= 0 ? num2[end2--] - '0' : 0;
            int x = x1 + x2 + next;

            //处理一下进位
            next = x / 10;
            x = x % 10;

            //尾插
            str += ('0' + x);
        }
        if(next == 1)
        {
            str += '1';
        }
        reverse(str.begin() , str.end());
        return str;
    }
};

3.Leetcode----344.反转字符串

3.1用双指针解决问题

class Solution {
public:
    void reverseString(vector<char>& s)
     {
        int n = s.size();
        for(int left = 0,right = n - 1;left < right;left ++ , right--)
        {
            swap(s[left] , s[right]);
        }
    }
};

4.Leetcode387--- 字符串中的第一个唯一字符

class Solution {
public:
    int firstUniqChar(string s)
     {
        //统计次数
        int count[26] = {0};
        for(auto ch : s)
        {
            count[ch-'a']++;
        }
        for(size_t i = 0;i < s.size();++i)
        {
            if(count[s[i] - 'a' ]== 1)
            {
                return i;
            }
        }
        return -1;
    }
};

5.Leetcode----125 验证回文串

class Solution {
public:
    bool isLetterOrNumber(char ch)
    {
        return (ch>='0'&&ch <='9')
        || (ch >='a'&& ch <='z')
        || (ch>='A' && ch <= 'Z');
    }
    bool isPalindrome(string s) 
    {
        for(auto& ch : s)
        {
            if(ch >='A' && ch <= 'Z')
            {
                ch += 32;
            }
        }
        int begin = 0 , end = s.size()-1;
        while(begin < end)
        {
            while(begin < end && !isLetterOrNumber(s[begin]))
            ++begin;
            while(begin < end && !isLetterOrNumber(s[end]))
            --end;
            if(s[begin] != s[end])
            {
                return false;
            }
            else
            {
                ++begin;
                --end;
            }
        }
        return true;
    }
};

6.Leetcode---43.字符串相乘

class Solution 

{

public:



  void MulItem(string &tmp, string &num1, char a)

  {

    int i = 0, sign=0;

    int mul = 0;

    while(i < num1.size())

    {

      mul = (num1[i]-'0') * (a-'0') + sign;

      if(mul >= 10)

      {

        sign = mul / 10;

        mul %= 10;

      }

      else

        sign = 0;

      tmp.push_back(mul+'0');

      i++;

    }    

    if(sign > 0)

      tmp.push_back(sign+'0'); 

  }   

  //对应为相加，sign进位采用引用传递

  int AddItem(int a, int b, int &sign)

  {

    int add = a+b+sign;

    if(add >= 10)

    {

      sign = 1;

      add -= 10;

    }

    else

      sign = 0;

    return add;

  }

  //错位相加

  void MoveAdd(string &result, string &tmp, int k)

  {

    int i, j;

    i = k;

    j = 0;

    int sign = 0;

    while(i<result.size() &&j<tmp.size())

    {

      result[i] = AddItem(result[i]-'0', tmp[j]-'0', sign) + '0';

      i++;

      j++;

    }

    while(i<result.size() && sign)

    {

      result[i] = AddItem(result[i]-'0', 0, sign)+'0';

      i++;

    }

    while(j < tmp.size())

    {

      int v = AddItem(0, tmp[j]-'0', sign);

      result.push_back(v+'0');

      j++;

    }

    if(sign)

      result.push_back(sign+'0');

  }



  string multiply(string num1, string num2) 

  {

  	 //先翻转数据，方便进位处理

     reverse(num1.begin(), num1.end());

     reverse(num2.begin(), num2.end());



     string tmp, result;

     for(int i=0; i<num2.size(); ++i)

     {

     	 //使用num2的每一个数据乘以num1

       MulItem(tmp, num1, num2[i]);

       //将乘得的结果进行错位相加

       MoveAdd(result, tmp, i);

       tmp.clear();

     }    



     while(result.size()!=1 && result.back()=='0')

      result.pop_back();



		 //翻转数据，恢复数据

     reverse(result.begin(), result.end());

     return result;

  }

};