From daf6e6b989bcf85ebc614c4a7c559cfdb2bc7839 Mon Sep 17 00:00:00 2001
From: liangjinglin <liangjinglin@hc-smart.com>
Date: Sun, 19 Jan 2025 14:51:08 +0800
Subject: [PATCH] =?UTF-8?q?2025-01-19=20=E6=96=90=E6=B3=A2=E9=82=A3?=
 =?UTF-8?q?=E5=A5=91=E6=9F=A5=E6=89=BE=E6=B3=95?=
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---
 .../java/Search/FibonacciSearch.java          | 81 +++++++++++++++++++
 1 file changed, 81 insertions(+)
 create mode 100644 src/dataANDcalc/java/Search/FibonacciSearch.java

diff --git a/src/dataANDcalc/java/Search/FibonacciSearch.java b/src/dataANDcalc/java/Search/FibonacciSearch.java
new file mode 100644
index 0000000..5f65fa7
--- /dev/null
+++ b/src/dataANDcalc/java/Search/FibonacciSearch.java
@@ -0,0 +1,81 @@
+package Search;
+
+import SortAlgori.MergeSort;
+
+import java.util.Arrays;
+
+public class FibonacciSearch {
+
+    private int[] arr;
+
+    // 斐波那契数列
+    private int[] fibonacci;
+
+    public FibonacciSearch(int[] arr){
+        this.arr = arr;
+        this.fibonacci = new int[20];
+        fibonacci[0] = 1;
+        fibonacci[1] = 1;
+        for (int i = 2; i < fibonacci.length; i++) {
+            fibonacci[i] = fibonacci[i-1] + fibonacci[i-2];
+        }
+    }
+
+    //{1,1,2,3,5,8,13,21,34,55}
+    //{1,8,10,89,1000,1234}
+    private int fibSearch(int key){
+        int low = 0;
+        int high = arr.length-1;
+        int k = 0;
+        int mid = 0;
+        // 让数组的长度去匹配斐波那契数列，直到找到数组长度大于斐波那契数组中的某个数为止
+        while(high > fibonacci[k]-1){
+            k++;
+        }
+        // 拓展原数组让原数组长度等于获取到的斐波那契数组的值f[k]
+        int[] temp = Arrays.copyOf(arr, fibonacci[k]);
+        // 拓展的部分用原数组的最后的元素进行填充
+        for(int i = high+1; i < temp.length; i++){
+            temp[i] = arr[high];
+        }
+
+        while(low <= high){
+            //获取斐波那契分割点的下标
+            //根据斐波那契数组的特性,F[K]=F[K-1]+F[K-2]可以得到F[K]-1=(F[K-1]-1)+(F[K-2]-1)+1
+            // 该表达式说明只要顺序表长度为F[K]-1则可以分割成F[K-1]-1和F[K-2]-1两段，所以mid=low+F[K-1]-1
+            mid = low + fibonacci[k-1] - 1;
+            if(key < temp[mid]){
+                high = mid - 1;
+                // 全部元素 = 前面元素 + 后面元素
+                //   f[k]  =  f[k-1] + f[k-2]
+                // 所以mid值前面的数列实际上就是f[k-1]
+                k--;
+            } else if (key > temp[mid]){
+                low = mid + 1;
+                // 全部元素 = 前面元素 + 后面元素
+                //   f[k]  =  f[k-1] + f[k-2]
+                // 所以mid值后面的数列实际上就是f[k-2]
+                k = k -2;
+            } else {
+                if (mid <= high){
+                    return mid;
+                } else {
+                    // 因为high以后的坐标是拓展队列
+                    return high;
+                }
+            }
+        }
+        return -1;
+    }
+
+    public static void main(String[] args) {
+        int[] arr = new int[]{1,8,10,89,1000,1234};
+        System.out.println("排序后的序列：");
+        FibonacciSearch fibonacciSearch = new FibonacciSearch(arr);
+        long startTime = System.currentTimeMillis();
+        int search = fibonacciSearch.fibSearch(1000);
+        long endTime = System.currentTimeMillis();
+        System.out.println("查找到的下标是：" + search);
+        System.out.println("查找所花费的时间：" + (endTime - startTime) + "ms");
+    }
+}