本题要求实现二分查找算法。
函数接口定义：
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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：
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2 |
输入样例2：
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3 26 78 233 31 |
输出样例2：
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1 |
二分查找算法图示：
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：
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