给出三个字符串:s1、s2、s3,推断s3是否由s1和s2交叉构成。
Yes
例子
比方 s1 = "aabcc" s2 = "dbbca"
- 当 s3 = "aadbbcbcac",返回 true.
- 当 s3 = "aadbbbaccc", 返回 false.
挑战
标签 Expand 要求时间复杂度为O(n^2)或者更好
相关题目 Expand
分析:两个字符串的问题,大部分都能够用dp[i][j]表示第一个字符串前i个字符第二个字符串前j个字符的匹配情况来解决
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()]; } };