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;}elsesign = 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;}elsesign = 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num1MulItem(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;}};
