Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.
Example：
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
解释下题目：
就是每次从s1或者s2提取一个字母，然后构造一个新的字符串，问给定的s3是不是其中之一。
1. DP
实际耗时：2ms
public boolean isInterleave(String s1, String s2, String s3) {
int m = s1.length();
int n = s2.length();
//长度都不同，肯定不是
if (s3.length() != m + n) {
return false;
}
boolean dp[][] = new boolean[m + 1][n + 1];
//初始化
dp[0][0] = true;
//初始化第一列
for (int row = 1; row <= m; row++) {
dp[row][0] = dp[row - 1][0] && (s1.charAt(row - 1) == s3.charAt(row - 1));
if (!dp[row][0]) {
break;
}
}
//初始化第一行
for (int col = 1; col <= n; col++) {
dp[0][col] = dp[0][col - 1] && (s2.charAt(col - 1) == s3.charAt(col - 1));
if (!dp[0][col]) {
break;
}
}
//填充剩余的
for (int row = 1; row <= m; row++) {
for (int col = 1; col <= n; col++) {
dp[row][col] = (dp[row - 1][col] && s1.charAt(row - 1) == s3.charAt(row + col - 1))
|| (dp[row][col - 1] && s2.charAt(col - 1) == s3.charAt(row + col - 1));
}
}
return dp[m][n];
}
思路有点“类似”机器人走格子的题目。
