交叉字符串

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代码：

class Solution { 

    public: 

        /**

         * Determine whether s3 is formed by interleaving of s1 and s2.

         * @param s1, s2, s3: As description.

         * @return: true of false.

         */ 

        bool isInterleave(string s1, string s2, string s3) { 

            // write your code here 

            if(s3.length()!=s1.length()+s2.length()) 

                return false; 

            if(s1.length()==0) 

                return s2==s3; 

            if(s2.length()==0) 

                return s1==s3; 

            vector<vector<bool> > dp(s1.length()+1,vector<bool>(s2.length()+1,false)); 

            dp[0][0] = true; 

            for(int i=1;i<=s1.length();i++) 

                dp[i][0] = dp[i-1][0]&&(s3[i-1]==s1[i-1]); 

            for(int i=1;i<=s2.length();i++) 

                dp[0][i] = dp[0][i-1]&&(s3[i-1]==s2[i-1]); 

            for(int i=1;i<=s1.length();i++) 

            { 

                for(int j=1;j<=s2.length();j++) 

                { 

                    int t = i+j; 

                    if(s1[i-1]==s3[t-1]) 

                        dp[i][j] = dp[i][j]||dp[i-1][j]; 

                    if(s2[j-1]==s3[t-1]) 

                        dp[i][j] = dp[i][j]||dp[i][j-1]; 

                } 

            } 

            return dp[s1.length()][s2.length()]; 

        } 

    };  

lintcode截图：

交叉字符串

标签：amp   des   als   rmi   nbsp   images   orm   and   ++   

原文地址：http://www.cnblogs.com/aly15109725486/p/7235686.html