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 tTo 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 tWe 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 aWe 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.
该题目用到了三维动态规划，到目前为止没有理解深入。public boolean isScramble(String s1, String s2) { if (s1.equals(s2)) { return true; } int len1 = s1.length(); int len2 = s2.length(); boolean[][][] scrambled = new boolean[len1][len2][len1 + 1]; for (int i = 0; i < len1; i++) { for (int j = 0; j < len2; j++) { scrambled[i][j][0] = true; scrambled[i][j][1] = (s1.charAt(i) == s2.charAt(j)); } } for (int i = len1 - 1; i >= 0; i--) { for (int j = len2 - 1; j >= 0; j--) { for (int n = 2; n <= Math.min(len1 - i, len2 - j ); n++) { for (int m = 1; m < n; m++) { scrambled[i][j][n] |= scrambled[i][j][m] && scrambled[i + m][j + m][n - m] || scrambled[i][j + n - m][m] && scrambled[i + m][j][n - m]; if(scrambled[i][j][n]) break; } } } } return scrambled[0][0][len1]; }
Scramble String,布布扣,bubuko.com
原文地址：http://blog.csdn.net/u010378705/article/details/35575479
