最长公共前缀
两种方法：两两比较或者一起比较
cpp
class Solution {
public:
//两两比较，每次取出两个字符串的最长公共前缀作为基准和下一个进行比较，时间复杂度O(N*M)
string longestCommonPrefix(vector<string>& strs)
{
string ans = strs[0];
for(int i = 1; i < strs.size(); i++)
{
int index = 0;
while(index < ans.size() && index < strs[i].size())
{
if(ans[index] == strs[i][index])
++index;
else
break;
}
string tmp(ans.begin(), ans.begin() + index);
ans = tmp;
}
return ans;
}
};
class Solution {
public:
所有字符串一起比较，当字符不同时，结束比较返回，时间复杂度也是O(M*N)
string longestCommonPrefix(vector<string>& strs)
{
if(strs.size() == 0) return "" ;
if(strs.size() == 1) return strs[0];
int index = 0;
int flag = 1;
while(flag && index < strs[0].size())
{
char ch = strs[0][index];
for(auto& str : strs)
{
if(str == "")
return "";
if(index >= str.size())
{
flag = 0;
break;
}
if(ch != str[index])
flag = 0;
}
if(flag == 1)
index++;
}
string ans(strs[0].begin(), strs[0].begin()+index);
return ans;
}
};最长回文字符串
中心扩展算法
cpp
class Solution {
public:
//中心扩展算法
//遍历，每次从当前元素向左右扩展比较
//注意奇数和偶数长度回文串的不同，所以比较起点不同
string longestPalindrome(string s) {
int n = s.size();
int len = 0, prev = 0, next = 0;
int anslen = 0, ansprev= 0, ansnext = 0;
for(int i = 0; i < n; i++)
{
//奇数处理
prev = i;
next = i;
while(prev >= 0 && next < n && s[prev] == s[next])
{
prev--;
next++;
}
len = next - prev - 1;
if(len > anslen)
{
anslen = len;
ansprev = prev;
ansnext = next;
}
//偶数处理
prev = i;
next = i + 1;
while(prev >= 0 && next < n && s[prev] == s[next])
{
prev--;
next++;
}
len = next - prev - 1;
if(len > anslen)
{
anslen = len;
ansprev = prev;
ansnext = next;
}
}
return s.substr(ansprev + 1, anslen);
}
};二进制求和
大数相加或者相乘等都需要将字符串分为一个一个的数值进行运算
cpp
class Solution {
public:
string addBinary(string a, string b) {
//先翻字符串
reverse(a.begin(), a.end());
reverse(b.begin(), b.end());
int p1 = 0, p2 = 0, len1 = a.size(), len2 = b.size();
int tmp = 0;
string ans;
while(p1 < len1 || p2 < len2)
{
int num1 = p1 < len1 ? a[p1] - '0' : 0;
int num2 = p2 < len2 ? b[p2] - '0' : 0;
int sum = num1 + num2 + tmp;
tmp = sum / 2;
sum %= 2;
ans += sum + '0';
p1++;
p2++;
}
if(tmp) ans += tmp + '0';
reverse(ans.begin(), ans.end());
return ans;
}
};字符串相乘
解法一：每位相乘之后求进位，相加结果
cpp
class Solution {
public:
string multiply(string num1, string num2) {
if(num1 == "0" || num2 == "0")
return "0";
int len1 = num1.size(), len2 = num2.size();
reverse(num1.begin(), num1.end());
reverse(num2.begin(), num2.end());
string ans(len1 + len2, '0');
for(int i = 0; i < len1; i++)
{
//进行相乘
string tmp(i, '0');
int b = 0;//进位
for(int j = 0; j < len2; j++)
{
int a = (num1[i] - '0') * (num2[j] - '0') + b;
b = a / 10;
a %= 10;
tmp += a + '0';
}
if(b)
tmp += b + '0';
b = 0;
//将算出的字符串加到ans中
int k = 0;
for(k = 0; k < tmp.size(); k++)
{
int a = ans[k] - '0' + tmp[k] - '0' + b;
b = a / 10;
a %= 10;
ans[k] = a + '0';
}
for(k; k < ans.size() && b; k++)
{
int a = ans[k] - '0' + b;
b = a / 10;
a %= 10;
ans[k] = a + '0';
}
}
while(ans.size() >= len1 + len2 && ans.back() == '0')
ans.pop_back();
reverse(ans.begin(), ans.end());
return ans;
}
};解法二：每一步相乘之后放入数组下标位置，最后求进位
cpp
class Solution {
public:
string multiply(string num1, string num2) {
if(num1 == "0" || num2 == "0")
return "0";
int len1 = num1.size(), len2 = num2.size();
reverse(num1.begin(), num1.end());
reverse(num2.begin(), num2.end());
vector<int> tmp(len1 + len2 - 1, 0);
//无进位相乘
for(int i = 0; i < len1; i++)
for(int j = 0; j < len2; j++)
tmp[i + j] += (num1[i] - '0') * (num2[j] - '0');
//处理进位
string ans;
int b = 0;
for(int i = 0; i < tmp.size(); i++)
{
int a = tmp[i] + b;
b = a / 10;
a %= 10;
ans += a + '0';
}
if(b) ans += b + '0';
reverse(ans.begin(), ans.end());
return ans;
}
};