算法专题：线性DP

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！！！

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。