LeetCode97—Interleaving String

LeetCode97—Interleaving String

原题

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.

分析

动态规划，$dp[i][j]$表示前$i$个字符的s1子串和前$j$个字符的s2子串的是否能组成s3前$i+j-1$个字符的子串。
现在讨论$dp[i][j]$的递推关系：
1. 当s3的第$i+j-1$个字符来自于s1，即$s3[i+j-1]==s1[i-1]$，那么当$dp[i-1][j]$为$true$时，$dp[i][j]$也一定为$true$。
2. 同理，当s3的第$i+j-1$个字符来自于s2，即$s3[i+j-1]==s2[j-1]$，那么当$dp[i][j-1]$为$true$时，$dp[i][j]$也一定为$true$
3. 最后返回值为$dp[s1.size()][s2.size()]$

动归方程：

$$dp[i][j]= \begin{cases} dp[i-1][j] &\mbox{if $s3[i+j-1]==s1[i-1]$，$dp[i-1][j]$ is true}\\ dp[i][j-1] &\mbox{if $s3[i+j-1]==s2[j-1]$，$dp[i][j-1]$ is true} \end{cases}$$

代码

根据方程写出代码：

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

                if ((s1[i - 1] == s3[i + j - 1])&&(dp[i-1][j]))
                {
                    dp[i][j] = dp[i - 1][j];
                }

                if ((s2[j - 1] == s3[i + j - 1])&&(dp[i][j-1]) )
                {
                    dp[i][j] = dp[i][j - 1];
                }

            }
        }
        return dp[len1][len2];
    }
};