From:https://leetcode.com/problems/scramble-string/
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.
思路:动态规划。
数组dp[k][i][j]表示长度为k-1,s中从i开始的长为k-1的串和t中从j开始长为k-1的串是否是scrambled。
对于任意k,i,j,只要存在一种情况,使得两个子串满足scrambled,那么dp[k][i][j]为true,否则为false。
对于任意k,i,j,scramble的情形有两种,情形一:
如果两个绿色满足scrambled,且两个黄色满足scrambled,那么dp[k][i][j]为true。
情形二:
也是绿色为scrambled且黄色为scrambled,有dp[k][i][j]为true。
说白了,scramble就是要左边和左边一致且右边和右边一致,或者左边和右边一致,右边和左边一致。
故要求:
对任意的k,i,j, 有
boolean can = false;
for(int m=1; m<k && !can; ++m) {
can = dp[m-1][i][j] && dp[k-m-1][i+m][j+m] || dp[m-1][i][j+k-m] && dp[k-m-1][i+m][j];
}
dp[k-1][i][j] = can;
下面是代码:
public class Solution {
public boolean isScramble(String s1, String s2) {
int n = s1.length();
if(n != s2.length()) {
return false;
}
if(0 == n) {
return true;
}
boolean[][][] dp = new boolean[n][n][n];
// dp[k][i][j] means length is k-1, s1 from i, s2 from j.
for(int i=0; i<n; ++i) {
for(int j=0; j<n; ++j) {
dp[0][i][j] = (s1.charAt(i) == s2.charAt(j));
}
}
for(int k=2; k<=n; ++k) {
for(int i=n-k; i>=0; --i) {
for(int j=n-k; j>=0; --j) {
boolean can = false;
for(int m=1; m<k && !can; ++m) {
can = dp[m-1][i][j] && dp[k-m-1][i+m][j+m] || dp[m-1][i][j+k-m] && dp[k-m-1][i+m][j];
}
dp[k-1][i][j] = can;
}
}
}
return dp[n-1][0][0];
}
}