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.
题解
s3能否由s1, s2中的字符,保留前后顺序构造而成。
一刷:
方法一,二维dynamic programming。用dp[i][j]表示s3的前i+j个prefix可否由s1中前i个字符和s2中的前j个字符构成,则dp[i][j] = (dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) || (dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1))
Time complexity: O(mn). dp array of size mn is filled.
Space complexity: O(m*n)
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s3.length() != s1.length() + s2.length()) {
return false;
}
boolean dp[][] = new boolean[s1.length() + 1][s2.length() + 1];
for (int i = 0; i <= s1.length(); i++) {
for (int j = 0; j <= s2.length(); j++) {
if (i == 0 && j == 0) {
dp[i][j] = true;
} else if (i == 0) {
dp[i][j] = dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
} else if (j == 0) {
dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
} else {
dp[i][j] = (dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) || (dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
}
}
}
return dp[s1.length()][s2.length()];
}
}
理论上,仅用到相邻状态时,都可以将二维数组转化为一维数组。注意,将上式中的dp[i - 1][j]替换为dp[i],表示上一层i的状态。
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s3.length() != s1.length() + s2.length()) {
return false;
}
boolean dp[] = new boolean[s2.length() + 1];
for (int i = 0; i <= s1.length(); i++) {
for (int j = 0; j <= s2.length(); j++) {
if (i == 0 && j == 0) {
dp[j] = true;
} else if (i == 0) {
dp[j] = dp[j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
} else if (j == 0) {
dp[j] = dp[j] && s1.charAt(i - 1) == s3.charAt(i + j - 1);
} else {
dp[j] = (dp[j] && s1.charAt(i - 1) == s3.charAt(i + j - 1)) || (dp[j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1));
}
}
}
return dp[s2.length()];
}
}
二刷
思路同上。