判断是否字符串重组 Scramble String @LeetCode
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思路:
1 递归:
简单的说,就是s1和s2是scramble的话,那么必然存在一个在s1上的长度l1,将s1分成s11和s12两段,同样有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。
如果要能通过OJ,还必须把字符串排序后进行剪枝
http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Ascramble-string%EF%BC%8C%E4%B8%89%E7%BB%B4%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/
2 DP:
使用了一个三维数组boolean result[len][len][len],其中第一维为子串的长度,第二维为s1的起始索引,第三维为s2的起始索引。
result[k][i][j]表示s1[i...i+k]是否可以由s2[j...j+k]变化得来。
http://www.blogjava.net/sandy/archive/2013/05/22/399605.html
package Level5;import java.util.Arrays;/** * Scramble String * * 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. * */public class S87 {public static void main(String[] args) {String s1 = "abab";String s2 = "bbaa";System.out.println(isScramble(s1, s2));System.out.println(isScrambleDP(s1, s2));}public static boolean isScramble(String s1, String s2) {if(s1.length() != s2.length()){return false;} if(s1.length()==1 && s2.length()==1){ return s1.charAt(0) == s2.charAt(0); } // 排序后可以通过 char[] s1ch = s1.toCharArray();char[] s2ch = s2.toCharArray();Arrays.sort(s1ch);Arrays.sort(s2ch);if(!new String(s1ch).equals(new String(s2ch))){return false;}for(int i=1; i<s1.length(); i++){// 至少分出一个字符出来String s11 = s1.substring(0, i);String s12 = s1.substring(i);String s21 = s2.substring(0, i);String s22 = s2.substring(i);//System.out.println(s1 + "-" + s2 + ": "+ s11 + ", " + s12 + ", " + s21 + ", " + s22);// 检测前半部是否匹配if(isScramble(s11, s21) && isScramble(s12, s22)){return true;}// 前半部不匹配,检测后半部是否匹配s21 = s2.substring(0, s2.length()-i);s22 = s2.substring(s2.length()-i);if(isScramble(s11, s22) && isScramble(s12, s21)){return true;}}return false; }public static boolean isScrambleDP(String s1, String s2) {int len = s1.length();if(len != s2.length()){return false;}if(len == 0){return true;}char[] c1 = s1.toCharArray();char[] c2 = s2.toCharArray();// canTransform 第一维为子串的长度delta,第二维为s1的起始索引,第三维为s2的起始索引// canTransform[k][i][j]表示s1[i...i+k]是否可以由s2[j...j+k]变化得来。boolean[][][] canT = new boolean[len][len][len];for(int i=0; i<len; i++){for(int j=0; j<len; j++){// 如果字符串总长度为1,则取决于唯一的字符是否想到canT[0][i][j] = c1[i] == c2[j];}}for(int k=2; k<=len; k++){// 子串的长度for(int i=len-k; i>=0; i--){// s1[i...i+k]for(int j=len-k; j>=0; j--){// s2[j...j+k]boolean canTransform = false;for(int m=1; m<k; m++){// 尝试以m为长度分割子串// canT[k][i][j]canTransform = (canT[m-1][i][j] && canT[k-m-1][i+m][j+m]) ||// 前前后后匹配 (canT[m-1][i][j+k-m] && canT[k-m-1][i+m][j]);// 前后后前匹配if(canTransform){break;}}canT[k-1][i][j] = canTransform;}}}return canT[len-1][0][0];}}
同样思路,换一种写法:
这其实是一道三维动态规划的题目,我们提出维护量res[i][j][n],其中i是s1的起始字符,j是s2的起始字符,而n是当前的字符串长度,res[i][j][len]表示的是以i和j分别为s1和s2起点的长度为len的字符串是不是互为scramble。
有了维护量我们接下来看看递推式,也就是怎么根据历史信息来得到res[i][j][len]。判断这个是不是满足,其实我们首先是把当前s1[i...i+len-1]字符串劈一刀分成两部分,然后分两种情况:第一种是左边和s2[j...j+len-1]左边部分是不是scramble,以及右边和s2[j...j+len-1]右边部分是不是scramble;第二种情况是左边和s2[j...j+len-1]右边部分是不是scramble,以及右边和s2[j...j+len-1]左边部分是不是scramble。如果以上两种情况有一种成立,说明s1[i...i+len-1]和s2[j...j+len-1]是scramble的。而对于判断这些左右部分是不是scramble我们是有历史信息的,因为长度小于n的所有情况我们都在前面求解过了(也就是长度是最外层循环)。
上面说的是劈一刀的情况,对于s1[i...i+len-1]我们有len-1种劈法,在这些劈法中只要有一种成立,那么两个串就是scramble的。
总结起来递推式是res[i][j][len] = || (res[i][j][k]&&res[i+k][j+k][len-k] || res[i][j+len-k][k]&&res[i+k][j][len-k]) 对于所有1<=k<len,也就是对于所有len-1种劈法的结果求或运算。因为信息都是计算过的,对于每种劈法只需要常量操作即可完成,因此求解递推式是需要O(len)(因为len-1种劈法)。
如此总时间复杂度因为是三维动态规划,需要三层循环,加上每一步需要线行时间求解递推式,所以是O(n^4)。虽然已经比较高了,但是至少不是指数量级的,动态规划还是有很大有事的,空间复杂度是O(n^3)。
public class Solution { public boolean isScramble(String s1, String s2) { int len = s1.length(); if(len != s2.length()) { return false; } boolean[][][] canScramble = new boolean[len][len][len+1]; // i,j with sublength for(int i=0; i<len; i++) { for(int j=0; j<len; j++) { canScramble[i][j][1] = s1.charAt(i) == s2.charAt(j); // substring start from i with length 1, compared with substring start from j with length 1 } } for(int sublen=2; sublen<=len; sublen++) { // end_pos = x+(sublen-1) <= len-1, so x <= len-sublen for(int i=0; i<=len-sublen; i++) { for(int j=0; j<=len-sublen; j++) { for(int p=1; p<sublen; p++) { // split position canScramble[i][j][sublen] |= (canScramble[i][j][p] && canScramble[i+p][j+p][sublen-p]) || (canScramble[i][j+sublen-p][p] && canScramble[i+p][j][sublen-p]); } } } } return canScramble[0][0][len]; }}
http://blog.csdn.net/linhuanmars/article/details/24506703
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