leetCode 87. Scramble String
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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.
大意是字符串局部镜像转换的判断
规律在于只要在s1的任意位置一分为二镜像交换位置能够变成s2,则返回true,否则返回false。当然s1一分为二后还要继续分,这时候用递归就好了。
为了加快速度,还需要剪枝,这里是把字符串转换为字符数组然后排序实现,用map计数也可以
public static boolean isScramble(String s1, String s2) { if(s1.equals(s2))return true; if(s1.length()!=s2.length())return false; char[]A = s1.toCharArray(); Arrays.sort(A); String temp1 = String.valueOf(A); A = s2.toCharArray(); Arrays.sort(A); String temp2 = String.valueOf(A); if(!temp1.equals(temp2)){ return false; } for(int i=1;i<s1.length();i++){ if(isScramble(s1.substring(0,i),s2.substring(0,i))&& isScramble(s1.substring(i),s2.substring(i)) || isScramble(s1.substring(0,i),s2.substring(s2.length()-i))&& isScramble(s1.substring(i),s2.substring(0,s2.length()-i))){ return true; } } return false; }
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