参考练习习题总集
10. 正则表达式匹配
第一道题就是困难题让我很难蚌,真是磨人啊。
class Solution {
public:
bool isMatch(string s, string p) {
int * * matrix=new int * [p.size()+1];
for (int i=0;i<p.size()+1;i++)
{
matrix[i]=new int [s.size()+1];
for (int j=0;j<s.size()+1;j++)
matrix[i][j]=0;
}
matrix[0][0]=1;
for (int i=1;i<p.size();i+=2)
{
if (p[i]=='*') matrix[i+1][0]=1;
else break;
}
for (int i=1;i<p.size()+1;i++)
for (int j=1;j<s.size()+1;j++)
if (s[j-1]==p[i-1] or p[i-1]=='.')
{
if (matrix[i-1][j-1]==1)
{
matrix[i][j]=1;
}
}
else if (p[i-1]=='*')
{
if (matrix[i-2][j]==1)
{
matrix[i][j]=1;continue;
}
if (matrix[i][j-1]==1 and (s[j-1]==p[i-2] or p[i-2]=='.'))
matrix[i][j]=1;
}
bool result=matrix[p.size()][s.size()];
for (int i=0;i<p.size()+1;i++)
delete [] matrix[i];
return result;
}
};
44. 通配符匹配
嗨嗨嗨,思想差不多的,又水一道困难题。
class Solution {
public:
bool isMatch(string s, string p) {
int len1=s.size(),len2=p.size();
int * * matrix=new int * [len2+1];
for (int i=0;i<len2+1;i++)
{
matrix[i]=new int [len1+1];
for (int j=0;j<len1+1;j++)
matrix[i][j]=0;
}
matrix[0][0]=1;
for (int i=0;i<len2;i++)
if (p[i]=='*') matrix[i+1][0]=1;
else break;
for (int i=1;i<len2+1;i++)
for (int j=1;j<len1+1;j++)
{
if (s[j-1]==p[i-1] or p[i-1]=='?')
{
if (matrix[i-1][j-1]==1)
matrix[i][j]=1;
}
else if (p[i-1]=='*')
{
if (matrix[i-1][j]==1)
{
matrix[i][j]=1;continue;
}
if (matrix[i-1][j-1]==1 or matrix[i][j-1]==1)
matrix[i][j]=1;
}
}
bool result=matrix[len2][len1];
for (int i=0;i<len2+1;i++)
delete [] matrix[i];
return result;
}
};
45. 跳跃游戏 II
记忆化搜索。这里的记忆化搜索本质就是线性DP。做这道题时真是气死我了,一开始明明记着长度为0的时候需要特判,但是后面做着做着就忘了加上去,交了一次又发现if (temp[i]!=0) return temp[i];
没有加上去,连着犯了两次低级错误。
class Solution {
public:
vector<int> lb;vector<int> temp;
int jump(vector<int>& nums) {
if (nums.size()==1) return 0;
lb=nums;temp.resize(nums.size(),0);
return dfs(0);
}
int dfs(int i)
{
if (temp[i]!=0) return temp[i];
if (i+lb[i]>=lb.size()-1) return 1;
int result=1e4-1;
for (int j=lb[i];j>=1;j--)
{
int result_temp=dfs(i+j)+1;
if (result_temp<result)
result=result_temp;
}
temp[i]=result;
return result;
}
};
53. 最大子数组和(LCR 161 连续天数的最高销售额)
前缀和
class Solution {
public:
int maxSubArray(vector<int>& nums) {
int * lb=new int [nums.size()+1];lb[0]=0;
for (int i=1;i<nums.size()+1;i++)
lb[i]=lb[i-1]+nums[i-1];
int min_value=0,result=INT_MIN;
for (int i=1;i<nums.size()+1;i++)
{
int result_temp=lb[i]-min_value;
if (result_temp>result) result=result_temp;
if (lb[i]<min_value) min_value=lb[i];
}
delete [] lb;
return result;
}
};
线性DP
class Solution {
public:
int maxSubArray(vector<int>& nums) {
for (int i=1;i<nums.size();i++)
if (nums[i-1]>0)
nums[i]+=nums[i-1];
int result=INT_MIN;
for (int i=0;i<nums.size();i++)
if (nums[i]>result)
result=nums[i];
return result;
}
};
91. 解码方法
class Solution {
public:
int numDecodings(string s) {
if (s[0]=='0') return 0;
int * lb=new int [s.size()+1];
lb[0]=0;lb[1]=1;
for (int i=2;i<s.size()+1;i++)
if (s[i-1]=='0')
{
if (s[i-2]=='1' or s[i-2]=='2')
lb[i]=max(lb[i-2],1);
else return 0;
}
else
{
lb[i]=lb[i-1];
if (s[i-2]=='1' or (s[i-2]=='2' and s[i-1]<='6'))
lb[i]+=max(lb[i-2],1);
}
int result=lb[s.size()];
delete [] lb;
return result;
}
};
97. 交错字符串
记忆化搜索
class Solution {
public:
string string1,string2,string3;vector<vector<int>> temp;
bool isInterleave(string s1, string s2, string s3) {
if (s1.size()+s2.size()!=s3.size()) return false;
string1=s1;string2=s2;string3=s3;temp.resize(s1.size()+1,vector<int> (s2.size()+1,0));
return func(0,0);
}
bool func(int l1,int l2)
{
if (l1==string1.size() and l2==string2.size()) return true;
if (temp[l1][l2]!=0) return temp[l1][l2]==1;
bool result=false;
if (l1<string1.size() and string1[l1]==string3[l1+l2])
result|=func(l1+1,l2);
if (l2<string2.size() and string2[l2]==string3[l1+l2])
result|=func(l1,l2+1);
temp[l1][l2]=result?1:-1;
return result;
}
};
线性DP
class Solution {
public:
bool isInterleave(string s1, string s2, string s3) {
if (s1.size()+s2.size()!=s3.size()) return false;
int * * matrix=new int * [s1.size()+1];
for (int i=0;i<s1.size()+1;i++)
{
matrix[i]=new int [s2.size()+1];
for (int j=0;j<s2.size()+1;j++)
matrix[i][j]=0;
}
queue<pair<int,int>> dl;dl.push(make_pair(0,0));
while (1)
{
queue<pair<int,int>> temp;
if (dl.front().first+1<=s1.size())
temp.push(make_pair(dl.front().first+1,dl.front().second));
while (!dl.empty())
{
int x=dl.front().first,y=dl.front().second;
if (x==0 and y==0) matrix[x][y]=1;
else if (x==0)
{
if (matrix[x][y-1]==1 and s2[y-1]==s3[y-1])
matrix[x][y]=1;
}
else if (y==0)
{
if (matrix[x-1][y]==1 and s1[x-1]==s3[x-1])
matrix[x][y]=1;
}
else
{
if ((matrix[x][y-1]==1 and s2[y-1]==s3[x+y-1]) or (matrix[x-1][y]==1 and s1[x-1]==s3[x+y-1]))
matrix[x][y]=1;
}
if (dl.front().second+1<=s2.size())
temp.push(make_pair(dl.front().first,dl.front().second+1));
dl.pop();
}
if (temp.empty()) break;
dl=temp;
}
bool result=matrix[s1.size()][s2.size()];
for (int i=0;i<s1.size()+1;i++)
delete [] matrix[i];
return result;
}
};
115. 不同的子序列
wtql,感觉神功大成了。
class Solution {
public:
int numDistinct(string s, string t) {
int mod=1e9+7;
int * * matrix=new int * [s.size()+1];
for (int i=0;i<s.size()+1;i++)
{
matrix[i]=new int [t.size()+1];
for (int j=0;j<t.size()+1;j++)
matrix[i][j]=0;
}
for (int i=1;i<s.size()+1;i++)
for (int j=1;j<t.size()+1;j++)
if (s[i-1]==t[j-1])
{
matrix[i][j]=(matrix[i][j]+matrix[i-1][j])%mod;
matrix[i][j]=(matrix[i][j]+matrix[i-1][j-1])%mod;
if (j==1) matrix[i][j]+=1;
}
else matrix[i][j]=matrix[i-1][j];
int result=matrix[s.size()][t.size()];
for (int i=0;i<s.size()+1;i++)
delete [] matrix[i];
return result%mod;
}
};
119. 杨辉三角 II
简单题!!!文章来源:https://www.toymoban.com/news/detail-825793.html
class Solution {
public:
vector<int> getRow(int rowIndex) {
vector<vector<int>> lb={{1}};
for (int i=1;i<=rowIndex;i++)
{
vector<int> temp;
temp.push_back(1);
for (int j=1;j<i;j++)
temp.push_back(lb[i-1][j-1]+lb[i-1][j]);
temp.push_back(1);
lb.push_back(temp);
}
return lb[rowIndex];
}
};
198. 打家劫舍(LCR 089 打家劫舍)
class Solution {
public:
int rob(vector<int>& nums) {
if (nums.size()==1) return nums[0];
if (nums.size()==2) return max(nums[0],nums[1]);
int * lb=new int [nums.size()+1];
lb[0]=0;lb[1]=nums[0];lb[2]=max(nums[0],nums[1]);
for (int i=3;i<nums.size()+1;i++)
lb[i]=max(lb[i-2],lb[i-3])+nums[i-1];
int result=max(lb[nums.size()],lb[nums.size()-1]);
delete [] lb;
return result;
}
};
213. 打家劫舍 II(LCR 090 打家劫舍 II)
class Solution {
public:
int rob(vector<int>& nums) {
if (nums.size()==1) return nums[0];
if (nums.size()==2) return max(nums[0],nums[1]);
int * lb=new int [nums.size()];
lb[0]=0;lb[1]=nums[1];lb[2]=max(nums[1],nums[2]);
for (int i=3;i<nums.size();i++)
lb[i]=max(lb[i-2],lb[i-3])+nums[i];
int result1=max(lb[nums.size()-2],lb[nums.size()-1]);
lb[0]=0;lb[1]=nums[0];lb[2]=max(nums[0],nums[1]);
for (int i=3;i<nums.size();i++)
lb[i]=max(lb[i-2],lb[i-3])+nums[i-1];
int result2=max(lb[nums.size()-2],lb[nums.size()-1]);
delete [] lb;
return max(result1,result2);
}
};
就做到这里吧。文章来源地址https://www.toymoban.com/news/detail-825793.html
到了这里,关于算法专题:线性DP的文章就介绍完了。如果您还想了解更多内容,请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章,希望大家以后多多支持TOY模板网!