回文字符串的判断代码

一个可以判断字符串是否回文Pallindrome的c程序

#include <stdio.h>
#include <stdbool.h>
#include <string.h>

bool isPalindromeNumber(const char *string);

int main(int argc, const char * argv[])
{

    // insert code here...
    printf("Begin!
");
    
    char *string = "abcccbaa";
    bool ret = isPalindromeNumber(string);
    
    if (ret) {
        printf("string is palindrome number.
");
    } else {
        printf("string is not palindrome number.
");
    }

    
    return 0;
}

bool isPalindromeNumber(const char *string)
{
    bool isPNumber = false;
    
    int i;
    size_t j;
    
    i = 0;
    j = strlen(string) - 1;
    
    if (strlen(string) < 3) {
        return false;
    }
    
    while ( j-i > 1) {

        if (string[i] == string[j]) {
            i++;
            j--;
        } else {
            isPNumber = false;
            break;
        }
        
        if (j - i <= 1) {
            isPNumber = true;
        }
    }
    
    return isPNumber;
}

 在这个基础上可以判断一个字符串中最长的回文长度

#include <stdio.h>
#include <stdbool.h>
#include <string.h>

bool isPalindromeNumber(const char *string);
bool isPalindrome(const char *string, int sp, int ep);
int max(int a, int b);
int getMaxPalindromeLength(const char *string, int sp, int ep);

int main(int argc, const char * argv[])
{

    // insert code here...
    printf("Begin!
");
    
    char *string = "abcccbfedppdefabcdef";
    bool ret01 = isPalindromeNumber(string);
    
    if (ret01) {
        printf("string is palindrome number.
");
    } else {
        printf("string is not palindrome number.
");
    }
    
    size_t endPoint = strlen(string) - 1;
    bool ret02 = isPalindrome(string, 0, (int)endPoint);
    if (ret02) {
        printf("string is palindrome number.
");
    } else {
        printf("string is not palindrome number.
");
    }
    
    int length = getMaxPalindromeLength(string, 0, (int)endPoint);
    printf("max palindrome length %d
", length);

    return 0;
}

bool isPalindromeNumber(const char *string)
{
    bool isPNumber = false;
    
    int i;
    size_t j;
    
    if (strlen(string) < 3) {
        return false;
    }
    
    i = 0;
    j = strlen(string) - 1;
    
    while ( j-i > 1) {

        if (string[i] == string[j]) {
            i++;
            j--;
        } else {
            isPNumber = false;
            break;
        }
        
        if (j - i <= 1) {
            isPNumber = true;
        }
    }
    
    return isPNumber;
}

bool isPalindrome(const char *string, int sp, int ep)
{
    bool ret = false;
    
    if ( ep + 1 > strlen(string) ) {
        return false;
    }
    if (ep - sp < 2) {
        return false;
    }
    while (ep - sp > 1) {
        if (string[sp] == string[ep]) {
            sp++;
            ep--;
        } else {
            return false;
        }
        if (ep - sp <= 1) {
            return true;
        }
    }
    return ret;
}

int getMaxPalindromeLength(const char *string, int sp, int ep)
{
    printf("current string : ");
    for (int i=sp; i<=ep; i++) {
        printf("%c",string[i]);
    }
    printf("
");
    
    int length = 0;
    if (ep - sp + 1 < 3) {
        return 0;
    }
    if (isPalindrome(string, sp, ep)) {
        length = ep - sp + 1;
    } else {
        int tmp01 = getMaxPalindromeLength(string, sp + 1, ep);
        int tmp02 = getMaxPalindromeLength(string, sp, ep - 1);
        length = max(tmp01, tmp02);
    }
    return length;
}

int max(int a, int b)
{
    return (a >= b)?a:b;
}

还有其他的解决方法，貌似用了动态规划。