[LeetCode] 97. Interleaving String

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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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dp[i][j] 表示s3[0:i+j] 是否可由 s1[0:i] 和 s2[0:j]构成。

dp[i][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]))

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class Solution {public:    bool isInterleave(string s1, string s2, string s3) {        int len1 = s1.length();        int len2 = s2.length();        int len3 = s3.length();        if (len1 + len2 != len3) return false;        vector<vector<bool>> dp(len1+1, vector<bool>(len2+1, false));        dp[0][0] = true;        for (int i=1; i<=len1; ++i) {            dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1]);        }        for (int j=1; j<=len2; ++j) {            dp[0][j] = dp[0][j-1] && (s2[j-1] == s3[j-1]);        }        for (int i=1; i<=len1; ++i) {            for (int j=1; j<=len2; ++j) {                dp[i][j] = dp[i][j] || (dp[i-1][j] && (s1[i-1] == s3[i+j-1]));                dp[i][j] = dp[i][j] || (dp[i][j-1] && (s2[j-1] == s3[i+j-1]));            }        }        return dp[len1][len2];    }};