题目:
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.
代码:
class Solution { public: bool isScramble(string s1, string s2) { const int n1 = s1.size(); const int n2 = s2.size(); if (n1!=n2) { return false; } const int n = n1; int alpha[26] = {0}; for ( int i=0; i<n; ++i ){ alpha[s1[i]-‘a‘]++; alpha[s2[i]-‘a‘]--; } for ( int i=0; i<26; ++i ){ if ( alpha[i]!=0 ) return false; } // terminal condition if ( n==1 ) return s1[0]==s2[0]; // recursive process for ( int i=1; i<n; ++i ){ //cout << s1 << "," << s2 << ":" << i << endl; if ( ( Solution::isScramble(s1.substr(0,i), s2.substr(0,i)) && Solution::isScramble(s1.substr(i,n-i), s2.substr(i,n-i)) ) || ( Solution::isScramble(s1.substr(0,i), s2.substr(n-i,i)) && Solution::isScramble(s1.substr(i,n-i), s2.substr(0,n-i)) ) ) { return true; } } return false; } };
tips:
这道题的题意自己并没有理解好,引用一个网上其他人的理解如下:
(http://www.blogjava.net/sandy/archive/2013/05/22/399605.html)
由于一个字符串有很多种二叉表示法,貌似很难判断两个字符串是否可以做这样的变换。
“对付复杂问题的方法是从简单的特例来思考,从而找出规律。
先考察简单情况:
字符串长度为1:很明显,两个字符串必须完全相同才可以。
字符串长度为2:当s1="ab", s2只有"ab"或者"ba"才可以。
对于任意长度的字符串,我们可以把字符串s1分为a1,b1两个部分,s2分为a2,b2两个部分,满足((a1~a2) && (b1~b2))或者 ((a1~b2) && (a1~b2))”
理解了题意,代码也就写出来了。
具体还有几个细节需要注意:
1. 为了剪枝并加快速度,做了如下几件事情:
a) 判断s1与s2的长度是否相等
b) 判断s1与s2的每个字符数量是否相等(这里由于是字母所以用一个定长数组alpha[26]表示:某个字母在s1中出现一次+1,在s2中出现一次-1;最终alpha的每个元素都是0则证明s1与s2的每个字符数量相等。扩展一下,如果字符不止26个字母,包含其他字符呢?可以用hashmap表示)
2. 设定终止条件:
如果s1和s2长度已经为1,无法再分割了,就直接比较即可。
3. 在递归传入参数的时候,用到了substr(begin, num):
a) begin代表切取的第一个字符下标,num代表截取几个字符
b) 注意每次传入isScramble的字符长度相等
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上述的做法类似记忆化搜索,网上还有一种动态规划的解法,也学习了吧。
(http://blog.csdn.net/linhuanmars/article/details/24506703)