C ++中的回文分割
假设我们有一个输入字符串,那么当该分区的每个子字符串都是回文时,该字符串的分区就是回文分区。在本节中,我们必须找到回文分割给定字符串所需的最小切割数。因此,如果字符串像“ababbbabbababa”,则最小分割为回文。这里需要3个切口。回文是:babbbab|b|阿巴巴
为了解决这个问题,我们将遵循以下步骤-
n:=长度
分别定义nxn阶的cut矩阵和pal矩阵
对于i:=0至n
pal[i,i]:=true和cut[i,i]:=0
对于2到n范围内的len,
j:=i+len–1
如果len=2,则
如果str[i]=str[j],则pal[i,j]:=true
否则当str[i]=str[j]和pal[i+1,j-1]≠0pal[i,j]时:=true
如果pal[i,j]为true,则cut[i,j]:=0
其他
cut[i,j]:=cut[i,j]和(cut[i,k]+cut[k+1,j+1]+1)的最小值
cut[i,j]:=∞
对于范围i至j-1的k
对于介于0到n–len之间的i,执行
返回割[0,n-1]
示例
让我们看下面的实现以更好地理解-
#include <iostream> #include <climits> using namespace std; int min (int a, int b){ return (a < b)? a : b; } int minPalPartion(string str){ int n = str.size(); int cut[n][n]; bool pal[n][n]; //true when palindrome present for i to j th element for (int i=0; i<n; i++){ pal[i][i] = true; //substring of length 1 is plaindrome cut[i][i] = 0; } for (int len=2; len<=n; len++){ for (int i=0; i<n-len+1; i++){//find all substrings of length len int j = i+len-1; // Set ending index if (len == 2) //for two character string pal[i][j] = (str[i] == str[j]); else //for string of more than two characters pal[i][j] = (str[i] == str[j]) && pal[i+1][j-1]; if (pal[i][j] == true) cut[i][j] = 0; else{ cut[i][j] = INT_MAX; //initially set as infinity for (int k=i; k<=j-1; k++) cut[i][j] = min(cut[i][j], cut[i][k] + cut[k+1][j]+1); } } } return cut[0][n-1]; } int main(){ string str= "ababbbabbababa"; cout << "Min cuts for Palindrome Partitioning is: "<<minPalPartion(str); }
输入项
ababbbabbababa
输出结果
Min cuts for Palindrome Partitioning is: 3