LeetCode OJ:Interleaving String
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Interleaving String
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.
dp[i][j]表示从s1中取i个元素,从s2中取j个元素,能否构成s3的前i+j个元素,所以有
dp[i][j]成立的条件有,当dp[i-1][j]为真,即s1中前i-1个元素,s2中前j个元素能组成s3的前i+j-1个元素时,s1[i-1]如果等于s3[i+j-1],
或者是当dp[i][j-1]为真,即s1中前i个元素,s2中前j-1个元素能组成s3的前i+j-1个元素时,s2[j-1]如果等于s3[i+j-1],dp[i][j]为真;
初始化时,i为0即只从s2中取元素,或者是j为0即只从s1中取元素
class Solution {public: bool isInterleave(string s1, string s2, string s3) { if(s1.size()+s2.size()!=s3.size())return false; vector<vector<bool>> dp(s1.size()+1); for(int i=0;i<=s1.size();i++)dp[i].resize(s2.size()+1); dp[0][0]=true; for(int i=1;i<=s1.size();i++) dp[i][0]=dp[i-1][0]&&s1[i-1]==s3[i-1]; for(int j=1;j<=s2.size();j++) dp[0][j]=dp[0][j-1]&&s2[j-1]==s3[j-1]; for(int i=1;i<=s1.size();i++) for(int j=1;j<=s2.size();j++) dp[i][j]=(dp[i-1][j]&&s1[i-1]==s3[i+j-1])||(dp[i][j-1]&&s2[j-1]==s3[i+j-1]); return dp[s1.size()][s2.size()]; }};
递归+滚动数组
class Solution {public: bool isInterleave(string s1, string s2, string s3) { if(s1.length()+s2.length()!=s3.length()) return false; if(s1.length()<s2.length()) return isInterleave(s2,s1,s3); vector<bool> f(s2.length()+1,true); for(size_t i=1;i<=s2.length();++i) f[i]=s2[i-1]==s3[i-1]&&f[0]; for(size_t i=1;i<=s1.length();++i){ f[0]=s1[i-1]==s3[i-1]&&f[0]; for(size_t j=1;j<=s2.length();++j) f[j]=(s1[i-1]==s3[i+j-1]&&f[j]) ||(s2[j-1]==s3[i+j-1]&&f[j-1]); } return f[s2.length()]; }};
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