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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.
思路：做过几道动态规划了，终于也有了点感觉。一次就AC了，好开心啊~
用dp[m][n]来存储s3[0 ~ m+n-1]是否是s1[0~m-1]与s2[0~n-1]交替组成的。注意，没有必要存s3的维度，因为s1[0~m-1]与s2[0~n-1]只能匹配s3的m+n个字符，即第三个维度是确定的。
dp[i][j] 只有在一下两种情况下为真
① dp[i-1][j] 为真，并且 s1[i-1] == s3[i+j-1]
② dp[i][j-1] 为真，并且 s2[j-1] == s3[i+j-1]
class Solution { public: bool isInterleave(string s1, string s2, string s3) { int len1 = s1.length(); int len2 = s2.length(); int len3 = s3.length(); if(len3 != len1 + len2) return false; vector<vector<bool>> dp(len1 + 1, vector<bool>(len2 + 1, false)); dp[0][0] = true; for(int i = 1; i < len1 + 1; i++) { dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1]); } for(int j = 1; j < len2 + 1; j++) { dp[0][j] = dp[0][j-1] && (s2[j-1] == s3[j-1]); } for(int i = 1; i < len1 + 1; i++) { for(int j = 1; j < len2 + 1; 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[len1][len2]; } };
【leetcode】 Interleaving String (hard)
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原文地址：http://www.cnblogs.com/dplearning/p/4180319.html
