[Leetcode] Interleaving String

weixin_33753003

于 2019-01-08 03:59:43 发布

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文章标签：数据结构与算法

本文介绍了一种使用动态规划解决字符串s3是否由s1和s2交错组成的算法。通过构建二维布尔数组，逐层判断s1和s2的字符是否能与s3形成匹配，最终返回s3是否为s1和s2的交错结果。

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Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

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

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

动态规划，类似于求最长公共子序列。

 1 class Solution {
 2 public:
 3     bool isInterleave(string s1, string s2, string s3) {
 4         if (s1.size() + s2.size() != s3.size()) {
 5             return false;
 6         }
 7             
 8         bool a[1000][1000];
 9         a[0][0] = true;
10         for (int i = 1; i <= s1.size(); ++i) {
11             a[i][0] = a[i-1][0] && (s1[i-1] == s3[i-1]);
12         }
13         for (int i = 1; i <= s2.size(); ++i) {
14             a[0][i] = a[0][i-1] && (s2[i-1] == s3[i-1]);
15         }
16         for (int i = 1; i <= s1.size(); ++i) {
17             for (int j = 1; j <= s2.size(); ++j) {
18                 a[i][j] = (a[i-1][j] && s1[i-1] == s3[i+j-1]) || (a[i][j-1] && s2[j-1] == s3[i+j-1]);
19             }
20         }
21         return a[s1.size()][s2.size()];
22     }
23 };