[LeetCode] 97. Interleaving String_ Hard tag: Dynamic Programming-优快云博客

本文深入探讨了如何使用动态规划解决字符串s3是否由s1和s2交错组成的问题，通过构建二维数组mem来记录子问题的解决方案，最终确定s3是否能由s1和s2交错生成。

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

这个题目利用dynamic programming，mem[l1 + 1][l2 + 1] #mem[i][j] means that whether the first i characters of s1 and the first j characters of s2 be able to make first i + j characters of s3.

　　mem[i][j] = (mem[i - 1][j] and s1[i - 1] == s3[i + j - 1]) # when the last character of s3 matches the last of s1

　　　　　　 or (mem[i][j - 1]) and s2[j - 1] == s3[i + j - 1]) # when the last character of s3 matches the last of s2

initial: mem[i][0] = s1[:i] == s3[:i]

　　　　 mem[0][j] = s2[:j] == s3[:j]

code

class Solution:
    def interLeaveString(self, s1, s2, s3):
            l1, l2, l3 = len(s1), len(s2), len(s3)
            if l1 + l2 != l3: return False
            mem = [[False] * (l2 + 1) for _ in range(l1 + 1)]
            for i in range(l1 + 1):
                mem[i][0] = s1[:i] == s3[:i]
            for j in range(l2 + 1):
                mem[0][j] = s2[:j] == s3[:j]
            for i in range(1, 1 + l1):
                for j in range(1, 1 + l2):
                    mem[i][j] = (mem[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or (mem[i][j - 1] and s2[j - 1] == s3[i + j - 1])
            return mem[l1][l2]