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

 1 #include <stdio.h>

 2 #include <stdbool.h>

 3 #include <string.h>

 4 

 5 bool isPalindromeNumber(const char *string);

 6 

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

 8 {

 9 

10     // insert code here...

11     printf("Begin!\n");

12     

13     char *string = "abcccbaa";

14     bool ret = isPalindromeNumber(string);

15     

16     if (ret) {

17         printf("string is palindrome number.\n");

18     } else {

19         printf("string is not palindrome number.\n");

20     }

21 

22     

23     return 0;

24 }

25 

26 bool isPalindromeNumber(const char *string)

27 {

28     bool isPNumber = false;

29     

30     int i;

31     size_t j;

32     

33     i = 0;

34     j = strlen(string) - 1;

35     

36     if (strlen(string) < 3) {

37         return false;

38     }

39     

40     while ( j-i > 1) {

41 

42         if (string[i] == string[j]) {

43             i++;

44             j--;

45         } else {

46             isPNumber = false;

47             break;

48         }

49         

50         if (j - i <= 1) {

51             isPNumber = true;

52         }

53     }

54     

55     return isPNumber;

56 }

判断是否回文的代码 Code

 在这个基础上可以判断一个字符串中最长的回文长度

  1 #include <stdio.h>

  2 #include <stdbool.h>

  3 #include <string.h>

  4 

  5 bool isPalindromeNumber(const char *string);

  6 bool isPalindrome(const char *string, int sp, int ep);

  7 int max(int a, int b);

  8 int getMaxPalindromeLength(const char *string, int sp, int ep);

  9 

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

 11 {

 12 

 13     // insert code here...

 14     printf("Begin!\n");

 15     

 16     char *string = "abcccbfedppdefabcdef";

 17     bool ret01 = isPalindromeNumber(string);

 18     

 19     if (ret01) {

 20         printf("string is palindrome number.\n");

 21     } else {

 22         printf("string is not palindrome number.\n");

 23     }

 24     

 25     size_t endPoint = strlen(string) - 1;

 26     bool ret02 = isPalindrome(string, 0, (int)endPoint);

 27     if (ret02) {

 28         printf("string is palindrome number.\n");

 29     } else {

 30         printf("string is not palindrome number.\n");

 31     }

 32     

 33     int length = getMaxPalindromeLength(string, 0, (int)endPoint);

 34     printf("max palindrome length %d\n", length);

 35 

 36     return 0;

 37 }

 38 

 39 bool isPalindromeNumber(const char *string)

 40 {

 41     bool isPNumber = false;

 42     

 43     int i;

 44     size_t j;

 45     

 46     if (strlen(string) < 3) {

 47         return false;

 48     }

 49     

 50     i = 0;

 51     j = strlen(string) - 1;

 52     

 53     while ( j-i > 1) {

 54 

 55         if (string[i] == string[j]) {

 56             i++;

 57             j--;

 58         } else {

 59             isPNumber = false;

 60             break;

 61         }

 62         

 63         if (j - i <= 1) {

 64             isPNumber = true;

 65         }

 66     }

 67     

 68     return isPNumber;

 69 }

 70 

 71 bool isPalindrome(const char *string, int sp, int ep)

 72 {

 73     bool ret = false;

 74     

 75     if ( ep + 1 > strlen(string) ) {

 76         return false;

 77     }

 78     if (ep - sp < 2) {

 79         return false;

 80     }

 81     while (ep - sp > 1) {

 82         if (string[sp] == string[ep]) {

 83             sp++;

 84             ep--;

 85         } else {

 86             return false;

 87         }

 88         if (ep - sp <= 1) {

 89             return true;

 90         }

 91     }

 92     return ret;

 93 }

 94 

 95 int getMaxPalindromeLength(const char *string, int sp, int ep)

 96 {

 97     printf("current string : ");

 98     for (int i=sp; i<=ep; i++) {

 99         printf("%c",string[i]);

100     }

101     printf("\n");

102     

103     int length = 0;

104     if (ep - sp + 1 < 3) {

105         return 0;

106     }

107     if (isPalindrome(string, sp, ep)) {

108         length = ep - sp + 1;

109     } else {

110         int tmp01 = getMaxPalindromeLength(string, sp + 1, ep);

111         int tmp02 = getMaxPalindromeLength(string, sp, ep - 1);

112         length = max(tmp01, tmp02);

113     }

114     return length;

115 }

116 

117 int max(int a, int b)

118 {

119     return (a >= b)?a:b;

120 }

获取最长回文的代码 Code

 

还有其他的解决方法，貌似用了动态规划。