Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 = "aabcc",
s2 = "dbbca",
When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.
思路
看s1和s2的首字母是否和s3的首字母相同，如果相同则分别进行计算。如果都不相同，返回false。
为了加快速度，使用带标记的递归，也可以用动规实现。
1 int r[100][100]; 2 void initialize(){ 3 int i,j; 4 for(i = 0; i < 100; i++){ 5 for(j = 0; j < 100; j++) 6 r[i][j] = -1; 7 } 8 r[0][0] = 1; 9 } 10 bool isInterleave2(string s1, string s2, string s3) { 11 // Note: The Solution object is instantiated only once and is reused by each test case. 12 int l1 = s1.length(), l2 = s2.length(), l3 = s3.length();
13 if(l1 + l2 != l3) 14 return false; 15 if(r[l1][l2] != -1) 16 return r[l1][l2]==1?true:false; 17 bool result = false; 18 if(l1 > 0 && s1[0] == s3[0]) 19 result = result||isInterleave2(s1.substr(1), s2, s3.substr(1)); 20 if(l2 > 0 && s2[0] == s3[0]) 21 result = result||isInterleave2(s1, s2.substr(1), s3.substr(1)); 22 if(result) 23 r[l1][l2] = 1; 24 else 25 r[l1][l2] = 0; 26 return result; 27 } 28 bool isInterleave(string s1, string s2, string s3) { 29 initialize(); 30 return isInterleave2(s1,s2,s3); 31 }
第二遍
bool isInterleave(string s1, string s2, string s3, int i1, int i2, int l1, int l2, vector<vector<int> > &result){ if(result[i1][i2] != -1) return result[i1][i2] == 1; bool tmp = false; if(i1 < l1 && s1[i1] == s3[i1+i2]) tmp = tmp || isInterleave(s1, s2, s3, i1+1, i2, l1, l2, result); if(i2 < l2 && s2[i2] == s3[i1+i2]) tmp = tmp || isInterleave(s1, s2, s3, i1, i2+1, l1, l2, result); result[i1][i2] = tmp; return tmp; } bool isInterleave(string s1, string s2, string s3) { int l1 = s1.length(), l2 = s2.length(), l3 = s3.length(); if(l1 + l2 != l3) return false; vector<vector<int> > result(l1+1, vector<int>(l2+1, -1)); result[l1][l2] = 1; return isInterleave(s1, s2, s3, 0, 0, l1, l2, result); }