给定三个字符串 s1, s2, s3, 验证 s3 是否是由 s1 和 s2 交错组成的。
示例 1:
代码语言：javascript复制输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
输出: true示例 2:
代码语言：javascript复制输入: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
输出: false解：动态规划，组成矩阵。看图，和前面几题都差不多。
代码语言：javascript复制class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
s1 = " " s1;
s2 = " " s2;
s3 = " " s3;
int len1 = s1.length();
int len2 = s2.length();
int len3 = s3.length();
if ((len1 len2) != len3) {
return false;
}
//dp[i][j]表示s1的0~i和s2的0~j能不能和s3的0~((i 1) (j 1)-1)组成交错字符
boolean[][] dp = new boolean[len1][len2];
//状态初始化
dp[0][0] = true;
for (int v = 1; v < len1; v ) {
dp[v][0] = dp[v - 1][0] && s1.charAt(v) == s3.charAt((v 1) (0 1) - 1);
}
for (int v = 1; v < len2; v ) {
dp[0][v] = dp[0][v - 1] && s2.charAt(v) == s3.charAt((0 1) (v 1) - 1);
}
//状态转移
for (int i = 1; i < len1; i ) {
for (int j = 1; j < len2; j ) {
dp[i][j] = (dp[i - 1][j] && s1.charAt(i) == s3.charAt((i 1) (j 1) - 1)) || (dp[i][j - 1] && s2.charAt(j) == s3.charAt((i 1) (j 1) - 1));
}
}
return dp[len1 - 1][len2 - 1];
}
}
