目录
1最长公共前缀
oj链接：最长公共前缀
cpp
// 解法1
class Solution
{
public:
string longestCommonPrefix(vector<string> &strs)
{
string ret = strs[0];
for (int i = 1; i < strs.size(); i++)
{
ret = FindComm(ret, strs[i]);
}
return ret;
}
string FindComm(string &s1, string &s2)
{
int i = 0;
while (i < min(s1.size(), s2.size()) && s1[i] == s2[i])
i++;
return s1.substr(0, i);
}
};
// 解法2
class Solution
{
public:
string longestCommonPrefix(vector<string> &strs)
{
for (int i = 0; i < strs[0].size(); i++) // 以第一个字符串为参照物进行比较
{
char tmp = strs[0][i];
for (int j = 1; j < strs.size(); j++)
{
if (i == strs[j].size() || strs[j][i] != tmp)
return strs[0].substr(0, i);
}
}
return strs[0];
}
};2最长回文子串
oj链接：最长回文子串
cpp
//动态规划版本 很重要！！
class Solution
{
public:
string longestPalindrome(string s)
{
int n = s.size();
vector<vector<bool>> dp(n, vector<bool>(n, false));
for (int i = n - 1; i >= 0; i--)
{
for (int j = i; j < n; j++)
{
if (s[i] == s[j])
{
if (i == j || i + 1 == j || dp[i + 1][j - 1])
dp[i][j] = true;
}
}
}
int begin = 0, len = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (dp[i][j])
{
if (j - i + 1 > len)
{
len = j - i + 1;
begin = i;
}
}
}
}
return s.substr(begin, len);
}
};
//中心扩展算法
class Solution
{
public:
string longestPalindrome(string s)
{
int n = s.size();
int begin = 0, len = 0;
for (int i = 0; i < n; i++)
{
// 奇数扩展
int left = i - 1, right = i + 1;
while (left >= 0 && right < n && s[left] == s[right])
{
left--; // 细节
right++;
}
int newlen = right - left - 1; // 细节
if (newlen > len)
{
begin = left + 1; // 细节
len = newlen;
}
// 偶数扩展
left = i, right = i + 1;
while (left >= 0 && right < n && s[left] == s[right])
{
left--;
right++;
}
newlen = right - left - 1;
if (newlen > len)
{
begin = left + 1;
len = newlen;
}
}
return s.substr(begin, len);
}
};3二进制求和
oj链接：二进制求和
cpp
class Solution
{
public:
string addBinary(string a, string b)
{
int cur1 = a.size() - 1, cur2 = b.size() - 1, t = 0;
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
oj链接：字符串相乘
cpp
// 解法1
class Solution
{
public:
string multiply(string num1, string num2)
{
if (num1 == "0" || num2 == "0")
return "0";
string ret = "0";
reverse(num1.begin(), num1.end());
reverse(num2.begin(), num2.end());
for (int i = 0; i < num1.size(); i++)
{
string tmp;
for (int a = 0; a < i; a++)
tmp += "0"; // 高位补0
int flag = 0;
for (int j = 0; j < num2.size(); j++)
{
int sum = (num1[i] - '0') * (num2[j] - '0') + flag;
flag = sum / 10;
tmp += (sum % 10) + '0';
}
if (flag != 0)
tmp += flag + '0';
// tmp类加到ret中
int cur1 = 0, cur2 = 0, t = 0;
string ret1;
while (cur1 < ret.size() || cur2 < tmp.size() || t)
{
if (cur1 < ret.size())
t += ret[cur1++] - '0';
if (cur2 < tmp.size())
t += tmp[cur2++] - '0';
ret1 += t % 10 + '0';
t /= 10;
}
ret = ret1;
}
// 记得要逆序回去
reverse(ret.begin(), ret.end());
return ret;
}
};
// 解法2
class Solution
{
public:
string multiply(string num1, string num2)
{
if (num1 == "0" || num2 == "0")
return "0";
int m = num1.size(), n = num2.size();
vector<int> tmp(m + n - 1, 0);
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
tmp[i + j] += (num1[i] - '0') * (num2[j] - '0');
}
}
string ret;
// tmp的数据写成string类型
int flag = 0;
for (int i = m + n - 2; i >= 0; i--)
{
int sum = tmp[i] + flag;
ret += sum % 10 + '0';
flag = sum / 10;
cout << ret << endl;
}
if (flag != 0)
ret += flag + '0';
reverse(ret.begin(), ret.end());
return ret;
}
};
class Solution
{
public:
// 无进位相加
string multiply(string num1, string num2)
{
if (num1 == "0" || num2 == "0")
return "0";
int n1 = num1.size(), n2 = num2.size();
// 为了能让数组与下标对应
reverse(num1.begin(), num1.end());
reverse(num2.begin(), num2.end());
vector<int> ret(n1 + n2 - 1);
for (int i = 0; i < n1; i++)
{
for (int j = 0; j < n2; j++)
{
ret[i + j] += ((num1[i] - '0') * (num2[j] - '0'));
}
}
string s = "0";
int cur = 0, t = 0;
string ans;
while (cur < n1 + n2 - 1 || t != 0)
{
if (cur < n1 + n2 - 1)
t += ret[cur++];
ans += t % 10 + '0';
t /= 10;
}
reverse(ans.begin(), ans.end());
return ans;
}
};以上便是全部内容，有问题欢迎在评论区指正：感谢观看！