97. Interleaving String

Description

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

Example

Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

思路

回溯法

• 最简单，也是最容易想到的思路。s1, s2, s3对应的下标分别是i, j, k。初始都指向对应的最后一个字符，
  然后判断相等的方式分别进行(i--,k--)或者(j--,k--)
• 缺点在于：时间复杂度太大。

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;

        return judge(s1, len1 - 1, s2, len2 - 1, s3, len3 - 1);
    }
private:
    bool judge(string &s1, int i, string& s2, int j, string &s3, int k)
    {
        if(i < 0 && j < 0 && k < 0) return true;
        if(k < 0)   return false;

        if(i >= 0 && j >= 0 && s1[i] == s2[j])
        {

            if(s3[k] != s1[i]) return false;
            if(judge(s1, i - 1, s2, j, s3, k - 1) || judge(s1, i, s2, j - 1, s3, k - 1))
                return true;
        }
        else if(i >= 0 && s1[i] == s3[k])
            return judge(s1, i - 1, s2, j, s3, k - 1);
        else if(j >= 0 && s2[j] == s3[k])
            return judge(s1, i, s2, j - 1, s3, k - 1);

        return false;
    }
};

动态规划

• 这其实也是一个很典型的动态规划问题。
• dp[i][j] 表示s3中长度为(i+j)的字符串是否由s1(0-i), s2(0-j)重构而成。若可以，取ture，
  否则为false
• dp[i][j] 中 i = 0 || j = 0, 表示有一个空串
• 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]));

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;

        vector<vector<bool>> dp(len1 + 1, vector<bool>(len2 + 1, false));

        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] && (s2[j - 1] == s3[i + j - 1]);
                }
                else if(j == 0)
                {
                    dp[i][j] = dp[i - 1][j] && (s1[i - 1] == s3[i + j - 1]);
                }
                else
                {
                    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];
    }
};