【題目描述】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.

【解題思路】這題最大的難度就是看懂題意。。。簡單的說，就是s1和s2是scramble的話，那么必然存在一個在s1上的長度l1，將s1分成s11和s12兩段，同樣有s21和s22。那么要么s11和s21是scramble的并且s12和s22是scramble的；要么s11和s22是scramble的并且s12和s21是scramble的。
解題方法是遞歸。觀察到要是兩個串相等，則必須滿足：1含有相同的字母；2當把兩個串分別拆成兩部分后，第一串的兩部分分別跟后一串的兩部分相比，只要比得上一次就相等。

【考查內容】樹，遞歸，動態(tài)規(guī)劃

class Solution { public:bool isScramble(string s1, string s2) {if (s1 == s2) return true; int A[26] = {0};for(int i = 0; i < s1.length(); i++) {A[s1[i]-'a']++;A[s2[i]-'a']--;}for (int i = 0; i < 26; i++) if (A[i] != 0) return false;for (int i = 1; i < s1.length (); ++i) {if (isScramble(s1.substr(0, i), s2.substr(0, i)) &&isScramble(s1.substr(i), s2.substr(i)))return true;if (isScramble(s1.substr(0, i), s2.substr(s1.length() - i)) &&isScramble(s1.substr(i), s2.substr(0, s1.length() - i)))return true;}return false;} };

總結

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