LeetCode 97 — Interleaving String （C++ Python）

题目：

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。设f[i][j]表示 s1[0,i] 和 s2[0,j]是否匹配 s3[0, i+j]。如果s1和s3的最后一个字符匹配，则f[i][j]=f[i-1][j]；如果s2和s3的最后一个字符匹配，则f[i][j]=f[i][j-1]。

C++实现：

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        int len1 = s1.size();
		int len2 = s2.size();
		
		if(s3.size() != len1 + len2)
		{
			return false;
		}

		vector<vector<bool> > f(len1 + 1, vector<bool>(len2 + 1, true));

		for(int i = 1; i <= len1; ++i)
		{
			f[i][0] = (s1[i - 1] == s3[i - 1]) && f[i - 1][0];
		}

		for(int j = 1; j <= len2; ++j)
		{
			f[0][j] = (s2[j - 1] == s3[j - 1]) && f[0][j - 1];
		}

		for(int i = 1; i <= len1; ++i)
			for(int j = 1; j <= len2; ++j)
			{ 
				f[i][j] = ((s1[i - 1] == s3[i + j - 1]) && f[i - 1][j]) ||
					((s2[j - 1] == s3[i + j - 1]) && f[i][j - 1]);	    

			}

		return f[len1][len2];  
    }
};

Python实现：

class Solution:
    # @return a boolean
    def isInterleave(self, s1, s2, s3):
        len1 = len(s1)
        len2 = len(s2)

        if len(s3) != len1 + len2:
            return False

        f = [[True] * (len2 + 1) for i in range(len1 + 1)] 

        for i in range(1, len1 + 1):
            f[i][0] = (s1[i - 1] == s3[i - 1]) and f[i - 1][0]
            
        for j in range(1, len2 + 1):
            f[0][j] = (s2[j - 1] == s3[j - 1]) and f[0][j - 1]

        for i in range(1, len1 + 1):
            for j in range(1, len2 + 1):
                f[i][j] = ((s1[i - 1] == s3[i + j - 1]) and f[i - 1][j]) or ((s2[j - 1] == s3[i + j - 1]) and f[i][j - 1])

        return f[len1][len2]

        感谢阅读！