判断一串字符中是否有回文串

#include<iostream>//C++版//判断是否包含回文字符串，如有，返回回文起始位置；如没有，返回字符串长度using namespace std;int find(char ch,char *str,size_t length) //查找str开始的字符串中第一个ch出现的位置,没找到返回-1{for(int i=0;i<length;i++){if(ch == str[i])return i;}return -1;}int isSym(char *str,size_t length)  //判断一个字符串是否全对称,0对称，-1不对称{for(int i=0;i<length/2;i++){if(str[i] != str[length-i-1])return -1;}return 0;}int isSymmetry(char *str,size_t length)//返回原字符串长度表示全部对称，-1表示无对称，其他值表示局部对称起始位置{if (isSym(str, length) == 0){return length;}int begin=0,end;char ch;for(int i=0;i<length;i++){ch = str[i];begin = end = i;while((begin = find(ch,str+end+1,length-end-1)) != -1){end = begin+end+1;if(isSym(str+i,end-i+1) == 0)return i;}}return -1;}int main(){char aa1[] = "123454321";char aa2[] = "123456547890";char aa3[] = "781234327891";char aa4[] = "954612313217891";char aa5[] ="3093166233266908312``";cout<<isSymmetry(aa1,sizeof(aa1)/sizeof(char)-1)<<endl;cout<<isSymmetry(aa2,sizeof(aa2)/sizeof(char)-1)<<endl;cout<<isSymmetry(aa3,sizeof(aa3)/sizeof(char)-1)<<endl;cout<<isSymmetry(aa4,sizeof(aa4)/sizeof(char)-1)<<endl;cout<<isSymmetry(aa5,sizeof(aa5)/sizeof(char)-1)<<endl;return 0;}

public class HuiWen{public static void main(String[]args){String str="987098321123211980890";HuiWen h=new HuiWen();int k=h.isOrNot(str);System.out.println(k);}public static int isOrNot(String str){char []A=str.toCharArray();int index;for(int i=0;i<A.length;i++){char t=A[i];if((index=find(A,t,i+1,A.length-1))!=-1){if(allOk(A,i,index)!=-1){return i;}}}return A.length;}public static int find(char A[],char x,int lo,int hi){int i=0;for(i=lo;i<=hi;i++){if(A[i]==x){return i;//找到输出下标}}return -1;//未找到，输出-1}public static int allOk(char A[],int lo,int hi){//判断整个字符串是否是回文int k=0;int i;for(i=lo;i<=hi;k++,i++){if(A[i]!=A[hi-k]){return -1;}}return 1;}}