[Leetcode][python]Interleaving String
2017-11-04 04:07
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题目大意
输入三个字符串s1、s2和s3,判断第三个字符串s3是否由前两个字符串s1和s2交替而成且不改变s1和s2中各个字符原有的相对顺序。解题思路
动态规划,思路容易理解。创建二位数组,然后[0][0]是True,首行首列则是假设其中一个字符串为空时,另一个字符串是否与目标字符串一一对应。
最后从[1][1]开始遍历,拿数字,与目标对应则那个方向的为True。
代码
class Solution(object): def isInterleave(self, s1, s2, s3): """ :type s1: str :type s2: str :type s3: str :rtype: bool """ if len(s1)+len(s2)!=len(s3): return False dp=[[False for i in range(len(s2)+1)] for j in range(len(s1)+1)] dp[0][0]=True for i in range(1, len(s1)+1): dp[i][0] = dp[i-1][0] and s3[i-1]==s1[i-1] # 边界:之前需要符合并且s1和s3对应位置相同 for i in range(1, len(s2)+1): dp[0][i] = dp[0][i-1] and s3[i-1]==s2[i-1] # 边界:之前需要符合并且s1和s3对应位置相同 for i in range(1, len(s1)+1): for j in range(1, len(s2)+1): # print i, j, s1[i-1], s3[i+j-1], s2[j-1], s3[i+j-1] dp[i][j] = (dp[i-1][j] and s1[i-1]==s3[i+j-1]) or (dp[i][j-1] and s2[j-1]==s3[i+j-1]) # for i in range(len(s1)+1): # print dp[i] return dp[-1][-1]
总结
还可以将而为动态规划压缩至一维,详见:https://shenjie1993.gitbooks.io/leetcode-python/097%20Interleaving%20String.html
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