leetCode 87.Scramble String （拼凑字符串） 解题思路和方法

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：这一天假设理解思想之后去做感觉不是非常难。可是在想到解法之前还是有一定难度。

本题解法为递归解法，动态规划解法没有掌握。

递归的思路和遍历字符串，切割成两部分，对照两部分能否scramble，只是本题要比較前前和前后两次。

详细代码例如以下：

public class Solution {
    public boolean isScramble(String s1, String s2) {
    	/**
    	 * 思想是递归，将字符串逐个的分为两串
    	 * 然后让两串的前前比較、后后比較，是则返回true
    	 * 不是着前后比較。是则返回true
    	 * 假设还不是。则分割的字符串位置+1
    	 */
    	
    	char[] c1 = s1.toCharArray();
    	char[] c2 = s2.toCharArray();

    	//调用函数推断
    	if(isScr(c1, c2, 0, c1.length, 0, c2.length)){
    		return true;
    	}
    	
		return false;
    }
	
    
    boolean isScr(char[] c1,char[] c2,int start1,int end1,int start2, int end2){
    	
    	//推断字符是否为空
    	if(end1 - start1 <= 0 && end2 - start2 <= 0){
    		return true;
    	}
    	//假设长度为1。则必须相等
    	if(end1 - start1 == 1 && end2 - start2 == 1){
    		return c1[start1] == c2[start2];
    	}
    	//长度不等返回false
    	if( end1 - start1 != end2 - start2){
    		return false;
    	}

    	int[] a = new int[128];
    	//每一个字符串的字符必须个数相等
    	for(int i = 0; i < end1 - start1; i++){
    		a[c1[i+start1]]++;
    		a[c2[i+start2]]--;
    	}

    	//不相等返回false
    	for(int i = 0; i < a.length; i++){
    		if(a[i] != 0){
    			return false;
    		}
    	}
    	//递归实现
    	for(int i = 1; i < end1 - start1; i++){
    		if(isScr(c1, c2, start1, start1+i, start2, start2+i) && isScr(c1, c2, start1+i, end1, start2+i, end2)){
    			return true;
    		}
    		
    		if((isScr(c1, c2, start1, start1+i, end2-i, end2) && isScr(c1, c2, start1+i, end1, start2, end2-i))){
    			return true;
    		}
    	}
    	return false;
    }
}