interpolationsearch.htm
来自「“常见程式演算”主要收集一些常见的程式练习题目」· HTM 代码 · 共 92 行
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<title>插补搜寻法</title>
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<h3><a href="http://caterpillar.onlyfun.net/GossipCN/index.html">From
Gossip@caterpillar</a></h3>
<h1><a href="AlgorithmGossip.htm">Algorithm Gossip: 插补搜寻法</a></h1>
<h2>说明</h2>
如果却搜寻的资料分布平均的话，可以使用插补（Interpolation）搜寻法来进行搜寻，在搜寻的对象大于500时，插补搜寻法会比 二分搜寻法 来的快速。<br>
<h2>解法</h2>
插补搜寻法是以资料分布的近似直线来作比例运算，以求出中间的索引并进行资料比对，如果取出的值小于要寻找的值，则提高下界，如果取出的值大于要寻找的
值，则降低下界，如此不断的减少搜寻的范围，所以其本原则与二分搜寻法是相同的，至于中间值的寻找是透过比例运算，如下所示，其中K是指定要寻找的对象，
而m则是可能的索引值： <br>
<div style="text-align: center;"><img style="width: 435px; height: 182px;" alt="插补搜寻" title="插补搜寻" src="images/interpolationSearch-1.jpg"></div>
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<h2> 实作</h2>
<ul>
<li> C
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<pre>#include <stdio.h> <br>#include <stdlib.h> <br>#include <time.h> <br>#define MAX 10 <br>#define SWAP(x,y) {int t; t = x; x = y; y = t;} <br><br>void quicksort(int[], int, int); <br>int intsrch(int[], int); <br><br>int main(void) { <br> int number[MAX] = {0}; <br> int i, find; <br><br> srand(time(NULL)); <br><br> for(i = 0; i < MAX; i++) { <br> number[i] = rand() % 100; <br> } <br><br> quicksort(number, 0, MAX-1); <br><br> printf("数列："); <br> for(i = 0; i < MAX; i++) <br> printf("%d ", number[i]); <br><br> printf("\n输入寻找对象："); <br> scanf("%d", &find); <br><br> if((i = intsrch(number, find)) >= 0) <br> printf("找到数字于索引 %d ", i); <br> else <br> printf("\n找不到指定数"); <br> <br> printf("\n"); <br><br> return 0; <br>} <br><br>int intsrch(int number[], int find) { <br> int low, mid, upper; <br><br> low = 0; <br> upper = MAX - 1; <br><br> while(low <= upper) { <br> mid = (upper-low)* <br> (find-number[low])/(number[upper]-number[low]) <br> + low; <br> if(mid < low || mid > upper) <br> return -1; <br><br> if(find < number[mid]) <br> upper = mid - 1; <br> else if(find > number[mid]) <br> low = mid + 1; <br> else <br> return mid; <br> } <br><br> return -1;<br>} <br><br>void quicksort(int number[], int left, int right) { <br> int i, j, k, s; <br><br> if(left < right) { <br> s = number[(left+right)/2]; <br> i = left - 1; <br> j = right + 1; <br><br> while(1) { <br> while(number[++i] < s) ; // 向右找 <br> while(number[--j] > s) ; // 向左找 <br> if(i >= j) <br> break; <br> SWAP(number[i], number[j]); <br> } <br><br> quicksort(number, left, i-1); // 对左边进行递回 <br> quicksort(number, j+1, right); // 对右边进行递回 <br> } <br>} <br></pre>
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<ul>
<li> Java
</li>
</ul>
<pre>public class InterpolationSearch {<br> public static int search(int[] number, int des) { <br> int low = 0; <br> int upper = number.length - 1; <br><br> while(low <= upper) { <br> int mid = (upper-low)* <br> (des-number[low])/(number[upper]-number[low]) <br> + low; <br> if(mid < low || mid > upper) <br> return -1; <br><br> if(des < number[mid]) <br> upper = mid - 1; <br> else if(des > number[mid]) <br> low = mid + 1; <br> else <br> return mid; <br> }<br><br> return -1;<br> }<br> <br> public static void main(String[] args) {<br> int[] number = {1, 4, 2, 6, 7, 3, 9, 8};<br> <br> QuickSort.sort(number);<br> <br> int find = InterpolationSearch.search(number, 3);<br> <br> if(find != -1) <br> System.out.println("找到数值于索引" + find); <br> else <br> System.out.println("找不到数值"); <br> } <br>}</pre>
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