SymmetricalString and AscendingNum

最新推荐文章于 2025-12-01 20:47:27 发布

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本文解析了两道华为面试题：一是判断字符串是否对称；二是使用递归判断整数数组是否升序排列。提供了多种实现方案及代码示例。

在论坛上遇到的两道华为面试题，试写如下：

1，判断一字符串是不是对称的，如：abccba。
2，用递归的方法 判断整数组a[N]是不是升序排列。

[1]

/* 判断一字符串是不是对称的，如：abccba 2010-3-15 wcdj */ #include <stdio.h> #include <assert.h> #include <string> #include <conio.h> bool SymmetricalString(const char* s); int main() { // test char* s1="abccba"; char* s2="abcdea"; if (SymmetricalString(s1)) { printf("s1 is symmetrical string./n"); } else { printf("s1 is not symmetrical string./n"); } if (SymmetricalString(s2)) { printf("s2 is symmetrical string./n"); } else { printf("s2 is not symmetrical string./n"); } getch(); return 0; } bool SymmetricalString(const char* s) { // check assert(s); // 先将原字符串保存在另一临时空间里 char stmp[128]={0}; strcpy(stmp,s); // 循环遍历比对 size_t nLen=strlen(stmp),i=nLen/2,j=0; while (i-->0) { if (stmp[j]!=stmp[nLen-j-1])// 分别比较左右边对应的元素 { return false; } else { ++j; } } return true; } /* 输出： s1 is symmetrical string. s2 is not symmetrical string. Press any key to continue */

完善一下SymmetricalString：

/* 判断一字符串是不是对称的，如：abccba 2010-3-15 wcdj */ #include <stdio.h> #include <assert.h> #include <string> #include <conio.h> bool SymmetricalString(const char* s); int main() { // test char* s1="abccba"; char* s2="abcdea"; if (SymmetricalString(s1)) { printf("s1 is symmetrical string./n"); } else { printf("s1 is not symmetrical string./n"); } if (SymmetricalString(s2)) { printf("s2 is symmetrical string./n"); } else { printf("s2 is not symmetrical string./n"); } getch(); return 0; } bool SymmetricalString(const char* s) { // check assert(s); // copy original string into a new temporal string char stmp[128]={0}; strcpy(stmp,s); size_t nLen=strlen(stmp); // check the length is even num if (!nLen || nLen%2)// 当nLen=0或者为奇数时返回false { return false; } // 循环遍历比对 for (int i=0;i!=nLen/2;++i) { if (stmp[i]!=stmp[nLen-i-1])// 分别比较左右边对应的元素 { return false; } } return true; } /* 输出： s1 is symmetrical string. s2 is not symmetrical string. Press any key to continue */

[2]

/* 用递归的方法判断整数组a[N]是不是升序排列 2010-3-15 wcdj */ #include <stdio.h> #include <assert.h> #include <conio.h> bool AscendingNum(const int* na); size_t nLen=0; int main() { // test int a1[]={1,3,5,6,8};// better int a2[]={3,5,1,6,8}; //int a1[]={'1','3','5','6','8'};// not good //int a2[]={'3','5','1','6','8'}; // calculate the length nLen=sizeof(a1)/sizeof(*a1); if (AscendingNum(a1)) { printf("a1 is ascending num/n"); } else { printf("a1 is not ascending num/n"); } // calculate the length again nLen=sizeof(a2)/sizeof(*a2); if (AscendingNum(a2)) { printf("a2 is ascending num/n"); } else { printf("a2 is not ascending num/n"); } getch(); return 0; } bool AscendingNum(const int* na) { // check assert(na); // 递归出口 if (nLen==1)// 仅剩最后一个元素没有比较 { return true; } if (*na<*(na+1)) { --nLen; AscendingNum(na+1); } else { return false; } } /* 输出： a1 is ascending num a2 is not ascending num Press any key to continue */

改进后的做法(借鉴huanmie_09朋友的做法)，也很巧妙：

#include<stdio.h> #define ARRAYLEN(_arr) (sizeof(_arr)/sizeof(*_arr)) int fun(int arr[], int n) { if(n == 1) { return 1; } if(n == 2) { return (arr[0] < arr[1]); } else { return (fun(arr+1, n-1) && arr[0] < arr[1]);// 巧妙之处 } } int main() { int arr1[] = {1, 2, 3, 4, 5}; int arr2[] = {1, 2, 5, 4, 2}; int flag1 = fun(arr1, ARRAYLEN(arr1)); if(flag1 != 0) { printf("yes/n"); } else { printf("no/n"); } int flag2 = fun(arr2, ARRAYLEN(arr2)); if(flag2 != 0) { printf("yes/n"); } else { printf("no/n"); } return 0; }

还有一种做法是从a[iLen - 1] >= a[iLen - 2]逆序比较：

BOOL IsIncreArray(int a[], int iLen) { if(iLen <= 0) return FALSE; else if (iLen == 1) return TRUE; else return IsIncreArray(a, iLen - 1) && (a[iLen - 1] >= a[iLen - 2]); }

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