LeetCode 97 [Interleaving String]

原题

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.

解题思路

• 典型序列型动态规划，Two Sequence DP
• 状态cache[i][j]表示，s1的前i个字符和s2的前j个字符是否能交叉构成s3的前i+j个字符
• 初始化：
• cache[0][0] = True 因为两个空字符串可以组成空字符串
• 边界情况是一个字符串为空，初始化只需判断另一个字符串和目标字符串前x为是否相等
• 递推关系 cache[i][j] = (s1[i] == s3[i+j] and cache[i-1][j]) or (s2[j] == s3[i+j] and cache[i][j-1])
• �最后说一个小小的简化程序的trick

if something > 0 and others < 0:
    x = True
# 直接可以写成
x = something > 0 and others < 0

完整代码

class Solution(object):
    def isInterleave(self, s1, s2, s3):
        """
        :type s1: str
        :type s2: str
        :type s3: str
        :rtype: bool
        """
        if len(s3) != len(s1) + len(s2):
            return False
            
        cache = [[False for i in range(len(s1) + 1)] for j in range(len(s2) + 1)]
        
        cache[0][0] = True
        for i in range(1, len(s1) + 1):
            if cache[0][i -1] and s1[i - 1] == s3[i - 1]:
                cache[0][i] = True
            else:
                break
                
        for j in range(1, len(s2) + 1):
            if cache[j - 1][0] and s2[j - 1] == s3[j - 1]:
                cache[j][0] = True
            else:
                break
                
        for j in range(1, len(s2) + 1):
            for i in range(1, len(s1) + 1):
                if (cache[j - 1][i] and s2[j - 1] == s3[j + i - 1]) or (cache[j][i - 1] and s1[i - 1] == s3[j + i - 1]):
                    cache[j][i] = True
                
        return cache[len(s2)][len(s1)]