97. Interleaving String
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问题描述:
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
Example 1:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac" Output: true
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc" Output: false
解题思路:
一开始我想用dfs,若当前s3的首字符能与s1的匹配,则匹配s1.substr(1) , s2, s3.substr(1)
若不成功则检查s3能否与s2匹配,若能则匹配s1, s2.substr(1), s3
超时了:)超时了!!!!!!!!!!!!我就知道hard没有这么简单啊朋友们TAT
遇到这样的字符匹配的题可以考虑dp
类似题目:72. Edit Distance
dp[i][j]表示s1前i个字符和s2前j个字符能否插入式构成 s3 前 i+j 个字符串
初始化:
当i = 0, j = 0 时,代表前0个字符串,所以一定为true
dp[i][0] = dp[i-1][0] && (s1[i-1] == s3[i-1])
dp[0][i] = dp[0][i-1] && (s2[i-1] == s3[i-1])
状态转移方程:
dp[i][j] = (dp[i-1][j] && s1[i-1] == s3[i-1] ) || (dp[i][j-1] && s1[j-1] == s3[j-1])
最后返回dp[len1][len2]
代码:
class Solution { public: bool isInterleave(string s1, string s2, string s3) { int len1 = s1.size(); int len2 = s2.size(); if(len1+len2 != s3.size()) 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 i = 1; i <=len2; i++){ dp[0][i] = dp[0][i-1] && (s2[i-1] == s3[i-1]); } for(int i = 1; i <= len1;i++){ for(int j = 1; j <= len2; 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]; } };
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