leetcode算法之字符串

1.最长公共前缀

最长公共前缀

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        //法一：两两比较
        string ret = strs[0];
        for(int i=1;i<strs.size();i++)
        {
            ret = findCommon(ret,strs[i]);
        }
        return ret;
    }
    string findCommon(string& s1,string& s2)
    {
        int i = 0;
        while(i<min(s1.size(),s2.size()) && s1[i]==s2[i]) i++;
        return s1.substr(0,i);
    }
};

class Solution {
public:
        string longestCommonPrefix(vector<string>& strs) {
        //法二：统一比较
        int n = strs[0].size();
        for(int i=0;i<n;i++)
        {
            char tmp = strs[0][i];
            for(int j = 1;j<strs.size();j++)
            {
                if(i == strs[j].size() || tmp != strs[j][i])
                {
                    return strs[0].substr(0,i);
                }
            }
        }
        return strs[0];
    }
};

2.最长回文子串

最长回文子串

class Solution {
public:
    string longestPalindrome(string s) {
        //使用中心扩展算法求解
        int begin = 0,len = 0,n = s.size();
        for(int i=0;i<n;i++)
        {
            //先做奇数长度的扩展
            int left = i,right = i;
            while(left>=0 && right<n && s[left] == s[right])
            {
                left--;
                right++;
            }
            if(right-left-1>len)
            {
                begin = left+1;
                len = right-left-1;
            }
            //再做偶数长度的扩展
            left = i,right = i+1;
            while(left>=0&&right<n&& s[left] == s[right])
            {
                left--;
                right++;
            }
            if(right-left-1>len)
            {
                begin = left+1;
                len = right - left-1;
            }
        }
        return s.substr(begin,len);
    }
};

3.二进制求和

二进制求和

class Solution {
public:
    string addBinary(string a, string b) {
        //模仿列竖式相加
        int t = 0;//表示进位
        int cur1 = a.size()-1;
        int cur2 = b.size()-1;
        string ret;
        while(cur1>=0 || cur2>=0 || t)
        {
            if(cur1>=0) t+=a[cur1--]-'0';
            if(cur2>=0) t+=b[cur2--]-'0';
            ret += t%2+'0';
            t /=2;
        }
        reverse(ret.begin(),ret.end());
        return ret;
    }
};

4.字符串相乘

字符串相乘

class Solution {
public:
    string multiply(string num1, string num2) {
        //高精度乘法
        //使用无进位相乘再相加，最后再处理进位
        //1.准备工作
        int m = num1.size(),n = num2.size();
        reverse(num1.begin(),num1.end());
        reverse(num2.begin(),num2.end());
        vector<int> tmp(m+n-1);
        //2.无进位相乘再相加
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                tmp[i+j] += (num2[i]-'0')*(num1[j]-'0');
            }
        }
        //3.处理进位
        int cur = 0,t = 0;
        string ret;
        while(cur<m+n-1 || t!=0)
        {
            if(cur<m+n-1) t+=tmp[cur++];
            ret += t%10+'0';
            t /= 10;
        }
        //4.处理前导零
        while(ret.size()>1 && ret.back() == '0') ret.pop_back();
        //5.注意输出结果顺序
        reverse(ret.begin(),ret.end());
        return ret;
    }
};