[Swift]LeetCode97. 交错字符串 | Interleaving String

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

Example 1:

Input: s1 = "aabcc",s2 = "dbbca",s3 = "aadbbcbcac"
Output: true

Example 2:

Input: s1 = "aabcc",s3 = "aadbbbaccc"
Output: false

给定三个字符串 s1,验证 s3 是否是由 s1 和 s2 交错组成的。

示例 1:

输入: s1 = "aabcc",s3 = "aadbbcbcac"
输出: true

示例 2:

输入: s1 = "aabcc",s3 = "aadbbbaccc"
输出: false

 1 // t = O(N*M),s = O(N*M)
 2 + (BOOL)isInterleaveWithS1: (Nsstring *)s1 s2: (Nsstring *)s2 s3: (Nsstring *)s3 {
 3     if (s1.length + s2.length != s3.length) {
 4         return false;
 5     }
 6     NSMutableArray<NSMutableArray<NSNumber *> *> *dp = [NSMutableArray arrayWithCapacity:s1.length];
 7     for (NSInteger i = 0; i < s1.length + 1; i++) {
 8         dp[i] = [NSMutableArray arrayWithCapacity:s2.length];
 9         for (NSInteger j = 0; j < s2.length + 1; j++) {
10             [dp[i] addobject:@(false)];
11         }
12     }
13     for (NSInteger i = 0; i < s1.length + 1; i++) {
14         for (NSInteger j = 0; j < s2.length + 1; j++) {
15             if (i == 0 && j == 0) {
16                 dp[i][j] = @(true);
17             } else if (i == 0) {
18                 dp[0][j] = @([dp[0][j-1] boolValue] && ([s2 characteratIndex:j-1] == [s3 characteratIndex:j-1]));
19             } else if (j == 0) {
20                 dp[i][0] = @([dp[i-1][0] boolValue] && ([s1 characteratIndex:i-1] == [s3 characteratIndex:i-1]));
21             } else {
22                 dp[i][j] = @(([dp[i][j-1] boolValue] && ([s2 characteratIndex:j-1] == [s3 characteratIndex:i+j-1])) || ([dp[i-1][j] boolValue] && ([s1 characteratIndex:i-1] == [s3 characteratIndex:i+j-1])));
23             }
24         }
25     }
26     return [dp[s1.length][s2.length] boolValue];
27 }