LeetCode: Interleaving String

LeetCode: Interleaving String

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.

地址：https://oj.leetcode.com/problems/interleaving-string/

算法：比较复杂的动态规划。dp[i][j][k]表示S3[0~i]是否可以由字符串S1[0~j-1]以及字符串S2[0~k-1]相交得到，并且i,j,k必须满足i+1=j+k。所以对于dp[0]，我们只需要初始化dp[0][1][0]以及dp[0][0][1]这两种，然后从dp[1]开始，从j可以取得的最大值开始，由于知道了i，j我们就可以求出k，这样我们就可以通过以下两种情况来求dp[i][j][k]的值，第一种状况，S3[i] 是否等于 S1[j-1]；第二种，S3[i]是否等于S2[k-1]。代码：

 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         if(s3.empty())
 7             return true;
 8         int len_s3 = s3.size();
 9         int len_s1 = s1.size();
10         int len_s2 = s2.size();
11         vector<vector<vector<bool> > > dp(len_s3,vector<vector<bool> >(len_s1+1,vector<bool>(len_s2+1)));
12         if(len_s1 > 0 && s3[0] == s1[0])
13             dp[0][1][0] = 1;
14         else if(len_s1 > 0)
15             dp[0][1][0] = 0;
16         if(len_s2 > 0 && s3[0] == s2[0])
17             dp[0][0][1] = 1;
18         else if(len_s2 > 0)
19             dp[0][0][1] = 0;
20         for(int i = 1; i < len_s3; ++i){
21             int most_s1 = min(i+1,len_s1);
22             int least_s1 = max(0,i-len_s2+1);
23             for(int j = most_s1; j >= least_s1 && i+1-j <= len_s2; --j){
24                 dp[i][j][i+1-j] = (j > 0 && dp[i-1][j-1][i+1-j] && s3[i] == s1[j-1]) || (i-j+1 > 0 && dp[i-1][j][i-j] && s3[i] == s2[i-j]);
25             }
26         }
27         return dp[len_s3-1][len_s1][len_s2];
28     }
29     int min(int a, int b){
30         return a>b ? b : a;
31     }
32     int max(int a, int b){
33         return a>b? a : b;
34     }
35 };