LintCode-交叉字符串-动态规划

描述

给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。

样例

比如 s1 = "aabcc" s2 = "dbbca"

- 当 s3 = "aadbbcbcac"，返回  true.

- 当 s3 = "aadbbbaccc"， 返回 false.

挑战

要求时间复杂度为O(n^2)或者更好

思路

State：
dp[i][j] 表示 A的前i个字符 与 B的前i个字符 能否交替组成 C的前 i+j的字符

Function：
dp[i][j] =
dp[i - 1][j] && (A[i-1] == C[i+j-1]) ||
dp[i][j - 1] && (B[j-1] == C[i+j-1])

Initialization:
dp[0][0] = true;
dp[i][0] = (A[0, 1, ..., i-1] == C[0, 1, ..., i-1])
dp[0][j] = (B[0, 1, ..., i-1] == C[0, 1, ..., i-1])

Answer:
dp[n][m]

代码

class Solution:
    """
    @param s1: A string
    @param s2: A string
    @param s3: A string
    @return: Determine whether s3 is formed by interleaving of s1 and s2
    """
    def isInterleave(self, s1, s2, s3):
        # write your code here
        n = len(s1)
        m = len(s2)
        if n == 0 and m == 0 and len(s3) == 0:
            return True
        if n+m != len(s3):
            return False
        dp = [[False for j in range(m+1)] for i in range(n+1)]
        for i in range(1, n+1):
            if s1[i-1] == s3[i-1]:
                dp[i][0] = True 
        for j in range(1, m+1):
            if s2[j-1] == s3[j-1]:
                dp[0][j] = True
        for i in range(1, n+1):
            for j in range(1, m+1):
                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[n][m]

题目来源

https://www.lintcode.com/problem/interleaving-string/description