LeetCode Scramble String

標籤：

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = "great":

    great   /      gr    eat / \    /  g   r  e   at           /           a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".

    rgeat   /      rg    eat / \    /  r   g  e   at           /           a   t

We say that "rgeat" is a scrambled string of "great".

Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".

    rgtae   /      rg    tae / \    /  r   g  ta  e       /       t   a

We say that "rgtae" is a scrambled string of "great".

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

題意：求s2是否是s1的拆分串。

思路：一種是遞迴搜尋，每次都是把一個串分成長度為i和len-i的兩個穿，再交叉比較；第二種是DP，設dp[len][i][j]表示的是s1第i個位置開始的長度為len的s2第j個位置開始的長度為len的是否滿足條件。

public class Solution {    public boolean isScramble(String s1, String s2) {        int len1 = s1.length();    int len2 = s2.length();    if (len1 != len2) return false;    if (len1 == 1) return s1.equals(s2);            char str1[] = s1.toCharArray();    char str2[] = s2.toCharArray();    Arrays.sort(str1);    Arrays.sort(str2);    if (Arrays.equals(str1, str2) == false) return false;        for (int i = 1; i < len1; i++) {    String s11 = s1.substring(0, i);    String s12 = s1.substring(i, len1);    String s21 = s2.substring(0, i);    String s22 = s2.substring(i, len1);    if (isScramble(s11, s21) && isScramble(s12, s22)) return true;    s21 = s2.substring(0, len1-i);    s22 = s2.substring(len1-i, len1);    if (isScramble(s11, s22) && isScramble(s12, s21)) return true;    }            return false;    }}

public class Solution {    public boolean isScramble(String s1, String s2) {      int len = s1.length();    if (len != s2.length()) return false;    if (len == 0) return false;        char c1[] = s1.toCharArray();    char c2[] = s2.toCharArray();        boolean dp[][][] = new boolean[len+1][len+1][len+1];    for (int i = 0; i < len; i++)    for (int j = 0; j < len; j++)    dp[1][i][j] = c1[i] == c2[j];        for (int k = 2; k <= len; k++)     for (int i = len-k; i >= 0; i--)    for (int j = len-k; j >= 0; j--) {    boolean flag = false;    for (int m = 1; m <= k && !flag; m++) {    flag = (dp[m][i][j] && dp[k-m][i+m][j+m]) ||    (dp[m][i][j+k-m] && dp[k-m][i+m][j]);    }    dp[k][i][j] = flag;    }            return dp[len][0][0];    }}

LeetCode Scramble String