97. Interleaving String

问题：

给定3个字符串，

由s1和s2构成s3。

若可以做到，返回true，否则返回false。

Example 1:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true

Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false

Example 3:
Input: s1 = "", s2 = "", s3 = ""
Output: true

Constraints:
0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1, s2, and s3 consist of lowercase English letters.

　　

example 1:

解法：DP

⚠️ 注意：本问题中，s1分成的n个子串，和s2分成的m个子串，

要求：|n-m|<1

其实，从s1中每次选一个or多个字符 or 从s2中选一个or多个字符，轮流进行选取，最终一定满足 |n-m|<1

1.状态：

• i：s1的第0～i个字符
• j：s2的第0～j个字符

2.选择：OR {

s3[i+j]

• ==s1[i]：选s1[i]   -> =dp[i-1][j]
• ==s2[j]：选s2[j] -> =dp[i][j-1]

}

3. dp[i][j]：

• 字符串s1选到第i个字符，s2选到第j个字符，是否能构成s3。

4.状态转移方程：

dp[i][j]= OR {

• s3[i+j]==s1[i] && dp[i-1][j]
• s3[i+j]==s2[i] && dp[i][j-1]

}

5.base：

• dp[0][0]=true
• dp[i][0] = s3[j]==s1[i] && dp[i-1][0]
• dp[0][j] = s3[j]==s2[j] && dp[0][j-1]

代码参考：

class Solution {
public:
    //dp[i][j][0]: s1[0~i] s2[0~j]
    //opt: OR :
    //case_1:  + s1[i-1]:
    //    cond:  s3[i+j-1]==s1[i-1]
    //       dp[i-1][j]
    //case_2:  + s2[j-1]:
    //    cond:  s3[i+j-1]==s2[j-1]
    //       dp[i][j-1]
    //base:
    //dp[0][0] = true
    //dp[0][x] = dp[0][x-1] && s3[x-1]==s2[x-1]
    //dp[x][0] = dp[x-1][0] && s3[x-1]==s1[x-1]
    bool isInterleave(string s1, string s2, string s3) {
        int len1 = s1.length();
        int len2 = s2.length();
        if(len1+len2!=s3.length()) return false;
        vector<vector<bool>> dp(len1+1, vector<bool>(len2+1, false));
        dp[0][0] = true;
        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] && s3[j-1]==s2[j-1]);
                else if(j==0) dp[i][j] = (dp[i-1][j] && s3[i-1]==s1[i-1]);
                else {
                    dp[i][j] = ((dp[i][j-1] && s3[i+j-1]==s2[j-1]) ||
                        (dp[i-1][j] && s3[i+j-1]==s1[i-1]));
                }
            }
        }
        return dp[len1][len2];
    }
};