【LeetCode with Python】 Interleaving String

博客域名：http://www.xnerv.wang
原题页面：https://oj.leetcode.com/problems/interleaving-string/
题目类型：动态规划
难度评价：★★★
本文地址：http://blog.csdn.net/nerv3x3/article/details/4258745

Given s1, s2, s3, find whether s3 is formed by the interleaving ofs1 ands2.

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

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

比较常规的一道字符串DP问题。注意检查传入参数，可以加速某些极端case的检测速度。此外还是用两个row代替一个完整的二维矩阵，节约空间消耗。

class Solution:    # @return a boolean    def isInterleave(self, s1, s2, s3):        len1 = len(s1)        len2 = len(s2)        len3 = len(s3)        if len1 + len2 != len3:            return False        if 0 == len1:            return s2 == s3        if 0 == len2:            return s1 == s3        row1 = [False for col in range(len2 + 1)]        row1[0] = True        for i in range(1, len2 + 1):            row1[i] = ( row1[i - 1] and (s2[i - 1] == s3[i - 1]))        row2 = [False for col in range(len2 + 1)]        for m in range(1, len1 + 1):            row2[0] = (row1[0] and (s1[m - 1] == s3[m - 1]))            for n in range(1, len2 + 1):                k = m + n - 1                # why and ? because [m, n] maybe come from up or left, one of the two is the right source.                row2[n] = (s1[m - 1] == s3[k] and row1[n]) or (s2[n - 1] == s3[k] and row2[n - 1])            row1, row2 = row2, row1        return row1[len2]