Interleaving String|leetcode题解
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这道题,我使用的是动态规划方法,对于字符串s2="dbbca",s1="aabcc",s3="aadbbcbcac"。设dp[i][j]代表s1[1..i]和s2[1...j]是否为s3[1...i+j]的交错串,则我们可以得出递归方程为dp[i][j]=(s3[i+j]==s1[i]&&dp[i-1][j])||(s3[i+j]==s2[j]&&dp[i][j-1]);即当s1和s2末尾字符和s3的末尾字符相等时,s3[1...i+j]是否为s1[1...i]和s2[1...j]的交错串,可以转化为s3[1...i+j-1]是否为s1[1...i-1]和s2[1...j]或者s1[1...i]和s2[1...j-1]交错串的判定。
下面为程序实现,在程序实现时,为了统一下标,在s1和s2前加了一个空格。
bool isInterleave(string s1, string s2, string s3) { int n1=s1.size(),n2=s2.size(),n3=s3.size(); if(n3!=n1+n2)return false; vector<vector<bool> >dp(n1+1,vector<bool>(n2+1,false)); s1=" "+s1; s2=" "+s2; s3=" "+s3; for(int i=0;i<=n1;i++){ for(int j=0;j<=n2;j++){ if(0==i&&0==j)dp[i][j]=1; else if(0==i)dp[i][j]=(s3[j]==s2[j]&&dp[0][j-1]); else if(0==j)dp[i][j]=(s3[i]==s1[i]&&dp[i-1][0]); else dp[i][j]=(s3[i+j]==s1[i]&&dp[i-1][j])||(s3[i+j]==s2[j]&&dp[i][j-1]); } } return dp[n1][n2]; }
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