712. Minimum ASCII Delete Sum for Two Strings

动态规划 
空间复杂度O(mn)
第一次提交
class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        //cout<<"s1: "<<s1<<endl;
        //cout<<"s2: "<<s2<<endl;
        int m=s1.size();
        int n=s2.size();
        vector<vector<int>> dp(m+1,vector<int>(n+1,0));
        for(int i=0;i<m;i++){
            dp[i+1][0]=dp[i][0]+s1[i];
            for(int j=0;j<n;j++){
                dp[0][j+1]=dp[0][j]+s2[j];
                if(s1[i]==s2[j])
                    dp[i+1][j+1]=dp[i][j];
                else
                    dp[i+1][j+1]=min(dp[i][j+1]+s1[i],dp[i+1][j]+s2[j]);
            }
        }
        /*
        for(int i=0;i<m+1;++i){
            for(int j=0;j<n+1;++j){
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }
        */
        return dp[m][n];
    }
};

 动态规划 空间复杂度 min(O(m),O(n))

class Solution {
public:
    int minimumDeleteSum(string s1, string s2) {
        //cout<<"s1: "<<s1<<endl;
        //cout<<"s2: "<<s2<<endl;
        //int m=s1.size();
        int m=s1.size();
        int n=s2.size();
        vector<int> dp(n+1,0);
        for(int j=0;j<n;++j)
            dp[j+1]=dp[j]+s2[j];
        for(int i=0;i<m;++i){
            int t1=dp[0];
            dp[0]=dp[0]+s1[i];
            for(int j=0;j<n;++j){
                int t2=dp[j+1];
                if(s1[i]==s2[j])
                    dp[j+1]=t1;
                else
                    dp[j+1]=min(dp[j+1]+s1[i],dp[j]+s2[j]);
                t1=t2;
            }
        }
        return dp[n];
    }
};

python 代码

class Solution(object):
    def minimumDeleteSum(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        m=len(s1)
        n=len(s2)
        dp=[[0 for _ in xrange(len(s2)+1)] for _ in xrange(len(s1)+1)]
        for j in xrange(1,len(s2)+1):
            dp[0][j]=dp[0][j-1]+ord(s2[j-1])
        for i in xrange(1,len(s1)+1):
            dp[i][0]=dp[i-1][0]+ord(s1[i-1])
            #print "dp[{}][0]: {}".format(i,dp[i][0])
            for j in xrange(1,len(s2)+1):
                if s1[i-1]==s2[j-1]:
                    dp[i][j]=dp[i-1][j-1]
                else:
                    dp[i][j]=min(dp[i-1][j]+ord(s1[i-1]),dp[i][j-1]+ord(s2[j-1]))
                    
                    
        #print dp
        
        return dp[-1][-1]