本题要求实现二分查找算法。
函数接口定义:
1 |
Position BinarySearch( List L, ElementType X ); |
其中List结构定义如下:
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typedef int Position; typedef struct LNode *List; struct LNode { ElementType Data[MAXSIZE]; Position Last; /* 保存线性表中最后一个元素的位置 */ }; |
L是用户传入的一个线性表,其中ElementType元素可以通过>、==、<进行比较,并且题目保证传入的数据是递增有序的。函数BinarySearch要查找X在Data中的位置,即数组下标(注意:元素从下标1开始存储)。找到则返回下标,否则返回一个特殊的失败标记NotFound。
裁判测试程序样例:
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#include <stdio.h> #include <stdlib.h> #define MAXSIZE 10 #define NotFound 0 typedef int ElementType; typedef int Position; typedef struct LNode *List; struct LNode { ElementType Data[MAXSIZE]; Position Last; /* 保存线性表中最后一个元素的位置 */ }; List ReadInput(); /* 裁判实现,细节不表。元素从下标1开始存储 */ Position BinarySearch( List L, ElementType X ); int main() { List L; ElementType X; Position P; L = ReadInput(); scanf("%d", &X); P = BinarySearch( L, X ); printf("%d\n", P); return 0; } /* 你的代码将被嵌在这里 */ |
输入样例1:
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5 12 31 55 89 101 31 |
输出样例1:
1 |
2 |
输入样例2:
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3 26 78 233 31 |
输出样例2:
1 |
二分查找算法图示:
![](https://zouzhongliang.com/wp-content/uploads/2019/06/二分法图示.jpg)
例:要查找20在数组中的位置,用二分法4次可查到。
二分查找算法代码:
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Position BinarySearch( List L, ElementType X ) { Position low = 1; Position high = L->Last; // Position pos; while (low<=high) { pos = (low + high) / 2; if (L->Data[pos] == X)return pos; else if (L->Data[pos] < X) { low = pos + 1; } else if (L->Data[pos] > X) { high = pos - 1; } } return NotFound; } |
二分法算法实际测试结果:
![](https://zouzhongliang.com/wp-content/uploads/2019/06/二分查找.jpg)
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