19.2.27 [LeetCode 97] Interleaving String

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

题意

给出s1，s2，s3，判断是否能从s3中抽取出若干个字符以原顺序组成字符串和s1相等，并且s3剩下的部分正好组成了s2

题解

自然而然想到的dp，很慢

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int l1=s1.length(),l2=s2.length(),l3=s3.length();
        vector<vector<vector<bool>>>dp(l1+1,vector<vector<bool>>(l2+1,vector<bool>(l3+1,false)));
        dp[0][0][0]=true;
        for(int i=0;i<=l1;i++)
            for(int j=0;j<=l2;j++){
                if(i>=1&&dp[i-1][j][i+j-1]&&s1[i-1]==s3[i+j-1])
                    dp[i][j][i+j]=true;
                else if(j>=1&&dp[i][j-1][i+j-1]&&s2[j-1]==s3[i+j-1])
                    dp[i][j][i+j]=true;
            }
        return dp[l1][l2][l3];
    }
};

好吧，我是傻x

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int l1=s1.length(),l2=s2.length(),l3=s3.length();
        if(l1+l2!=l3)return false;
        vector<vector<bool>>dp(l1+1,vector<bool>(l3+1,false));
        dp[0][0]=true;
        for(int i=0;i<=l1;i++)
            for(int j=0;j<=l2;j++){
                if(i>=1&&dp[i-1][i+j-1]&&s1[i-1]==s3[i+j-1])
                    dp[i][i+j]=true;
                else if(j>=1&&dp[i][i+j-1]&&s2[j-1]==s3[i+j-1])
                    dp[i][i+j]=true;
            }
        return dp[l1][l3];
    }
};