LeetCode:Interleaving String

Interleaving String

Total Accepted: 49969 Total Submissions: 220816 Difficulty: Hard

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.

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 Dynamic Programming String

思路：

s1 = "aabcc";

s2 = "dbbca";

s3 = "aadbbcbcac";

1）如果i==0 && j==0,dp[0][0] = true；

2）如果i==0,判断s2[j]==s3[i+j] 并且 dp[i][j-1]==true，则dp[i][j] = true；

3）如果j==0,与2）类似；

4）如果i != 0 && j != 0，判断(dp[i][j-1] && s2[j]==s3[i+j]) || (dp[i-1][j] && s1[i]==s3[i+j])；

5）dp[s1.length()][s2.length()]即为结果。

过程如下图所示：

j0 1  2  3  4  5  i 
s2dbbca 0 s1TFFFFF 1 aTFFFFF 2 aTTTTTF 3 bFTTFTF 4 cFFTTTT 5 cFFFTFT

c++ code:

class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        int len1 = s1.length();        int len2 = s2.length();        bool dp[len1+1][len2+1];                for(int i=0;i<=len1;i++) {            for(int j=0;j<=len2;j++) {                if(i==0 && j==0)                    dp[i][j] = true;                else if(i==0)                    dp[i][j] = (dp[i][j-1] && s2[j-1]==s3[i+j-1]);                else if(j==0)                    dp[i][j] = (dp[i-1][j] && s1[i-1]==s3[i+j-1]);                else                    dp[i][j] = (dp[i][j-1] && s2[j-1]==s3[i+j-1])||(dp[i-1][j] && s1[i-1]==s3[i+j-1]);            }        }        return dp[len1][len2];    }};