旋轉數組的最小數字（C++ 和 Python 實現）

（說明：本博客中的題目、題目詳細說明及參考代碼均摘自 “何海濤《劍指Offer：名企面試官精講典型編程題》2012年”）

題目

　　把一個數組最開始的若干個元素搬到數組的末尾，我們稱之為數組的旋轉。輸入一個遞增排序的數組的一個旋轉，輸出旋轉數組的最小元素。例如數組 {3, 4, 5, 1, 2} 為 {1, 2, 3, 4, 5} 的一個旋轉，該數組的最小值為 1 。

 

算法設計思想

1. 暴力查找（Bruteforce Search）：把旋轉數組從前到后遍歷一遍，其時間復雜度為 O(n)。很明顯，這種思想非常直接粗暴，沒有用到旋轉數組的性質。

2. 二分查找（Binary Search）：每次查找都把旋轉數組平均分成兩部分，通過比較當前旋轉數組兩端點和中間點的值，判斷最小值在數組的哪一部分，從而達到縮小搜索范圍的目的。其中，兩端點為當前的旋轉數組索引最小和索引最大的元素，中間點為這兩部分子數組的連結元素，也可以看做為軸樞點（pivot point），這里取旋轉數組的最小索引和最大索引的算術平均值（向下取整）作為索引，其對應的元素作為中間點。具體過程，如圖 2.10 所示。

需要注意，當旋轉數組的兩端點的值都與中間點的值相等時，因為無法判斷最小值在哪一部分，因此需要采用順序查找方法，即暴力查找。其示例如圖 2.11 所示。

 

C++ 實現

#include <stdio.h> #include <exception> #include <iostream>

// Declare a parameter exception
class ParamException: public std::exception { virtual const char* what() const throw() { return "Error: input parameters exception!"; } }; // find minimum in data[staIndex..endIndex]
int findMinInRotatedArray(int data[], int staIndex, int endIndex) { if (staIndex > endIndex || data == NULL) { throw ParamException();
 } int minimum = data[staIndex]; if (data[staIndex] >= data[endIndex] && staIndex < endIndex - 1) // 易錯點 { // Use Binary Search
        int midIndex = (staIndex + endIndex) / 2; if (data[midIndex] > data[staIndex]) minimum = findMinInRotatedArray(data, midIndex, endIndex); else if (data[midIndex] < data[endIndex]) minimum = findMinInRotatedArray(data, staIndex, midIndex); else { // Find the minimum sequentially
            for (int i = staIndex+1; i <= endIndex; ++i) if (minimum > data[i]) minimum = data[i]; } } else if (staIndex == (endIndex - 1)) { minimum = data[staIndex] > data[endIndex]? data[endIndex]:data[staIndex]; } return minimum; } void unitest() { int arr1[] = {3, 4, 5, 1, 2}; int arr2[] = {1, 0, 1, 1, 1}; int arr3[] = {1, 1, 1, 0, 1}; int arr4[] = {1, 2, 3, 4, 5}; printf("The minimum of the rotated array {3, 4, 5, 1, 2} is %d.\n", findMinInRotatedArray(arr1, 0, 4)); printf("The minimum of the rotated array {1, 0, 1, 1, 1} is %d.\n", findMinInRotatedArray(arr2, 0, 4)); printf("The minimum of the rotated array {1, 1, 1, 0, 1} is %d.\n", findMinInRotatedArray(arr3, 0, 4)); printf("The minimum of the rotated array {1, 2, 3, 4, 5} is %d.\n", findMinInRotatedArray(arr4, 0, 4)); // Test index parameter exception
    try { findMinInRotatedArray(arr3, 5, 4); } catch(std::exception& e) { std::cout << e.what() << std::endl; }; } int main(void) { unitest(); return 0; }

 

Python 實現

#!/usr/bin/python # -*- coding: utf8 -*-

# Define ParamError Exception
class ParamError(Exception): def __init__(self, value): self.value = value def __str__(self): return repr(self.value) # Find the minimum in rotated array 
def min_in_rotated_array(data, length): if data is None or length <= 0: raise ParamError("Error: input parameters exception!") # Index initialization
    sta, mid, end = 0, 0, length-1
    # Ensure this requisite before binary search
    while data[sta] >= data[end]: if end - sta == 1: mid = end break
        # Get the middle index
        mid = (sta + end) / 2
        # Find the minimum in order
        if (data[sta] == data[mid]) and (data[mid] == data[end]): minimum = data[sta] for i in range(sta+1, end+1): if minimum > data[i]: minimum = data[i] return minimum if data[sta] <= data[mid]: sta = mid elif data[end] >= data[mid]: end = mid return data[mid] def unitest(): arr1 = [3, 4, 5, 1, 2] arr2 = [1, 0, 1, 1, 1] arr3 = [1, 1, 1, 0, 1] arr4 = [1, 2, 3, 4, 5] print("The minimum of the rotated array [3, 4, 5, 1, 2] is %d." % min_in_rotated_array(arr1, 5)); print("The minimum of the rotated array [1, 0, 1, 1, 1] is %d." % min_in_rotated_array(arr2, 5)); print("The minimum of the rotated array [1, 1, 1, 0, 1] is %d." % min_in_rotated_array(arr3, 5)); print("The minimum of the rotated array [1, 2, 3, 4, 5] is %d." % min_in_rotated_array(arr4, 5)); try: min_in_rotated_array(arr1, -2) except Exception, e: print "\nFunction call: min_in_rotated_array(arr1, -2)"
        print e if __name__ == '__main__': unitest()

 

參考代碼

1. targetver.h

#pragma once

// The following macros define the minimum required platform. The minimum required platform // is the earliest version of Windows, Internet Explorer etc. that has the necessary features to run // your application. The macros work by enabling all features available on platform versions up to and // including the version specified. // Modify the following defines if you have to target a platform prior to the ones specified below. // Refer to MSDN for the latest info on corresponding values for different platforms.
#ifndef _WIN32_WINNT            // Specifies that the minimum required platform is Windows Vista.
#define _WIN32_WINNT 0x0600     // Change this to the appropriate value to target other versions of Windows.
#endif

2. stdafx.h

// stdafx.h : include file for standard system include files, // or project specific include files that are used frequently, but // are changed infrequently // 
#pragma once #include "targetver.h" #include <stdio.h> #include <tchar.h>



// TODO: reference additional headers your program requires here

3. stdafx.cpp

// stdafx.cpp : source file that includes just the standard includes // MinNumberInRotatedArray.pch will be the pre-compiled header // stdafx.obj will contain the pre-compiled type information
 #include "stdafx.h"

// TODO: reference any additional headers you need in STDAFX.H // and not in this file

4. MinNumberInRotatedArray.cpp

// MinNumberInRotatedArray.cpp : Defines the entry point for the console application. //

// 《劍指Offer——名企面試官精講典型編程題》代碼 // 著作權所有者：何海濤
 #include "stdafx.h" #include<exception>

int MinInOrder(int* numbers, int index1, int index2); int Min(int* numbers, int length) { if(numbers == NULL || length <= 0) throw new std::exception("Invalid parameters"); int index1 = 0; int index2 = length - 1; int indexMid = index1; while(numbers[index1] >= numbers[index2]) { // 如果index1和index2指向相鄰的兩個數， // 則index1指向第一個遞增子數組的最后一個數字， // index2指向第二個子數組的第一個數字，也就是數組中的最小數字
        if(index2 - index1 == 1) { indexMid = index2; break; } // 如果下標為index1、index2和indexMid指向的三個數字相等， // 則只能順序查找
        indexMid = (index1 + index2) / 2; if(numbers[index1] == numbers[index2] && numbers[indexMid] == numbers[index1]) return MinInOrder(numbers, index1, index2); // 縮小查找范圍
        if(numbers[indexMid] >= numbers[index1]) index1 = indexMid; else if(numbers[indexMid] <= numbers[index2]) index2 = indexMid; } return numbers[indexMid]; } int MinInOrder(int* numbers, int index1, int index2) { int result = numbers[index1]; for(int i = index1 + 1; i <= index2; ++i) { if(result > numbers[i]) result = numbers[i]; } return result; } // ====================測試代碼====================
void Test(int* numbers, int length, int expected) { int result = 0; try { result = Min(numbers, length); for(int i = 0; i < length; ++i) printf("%d ", numbers[i]); if(result == expected) printf("\tpassed\n"); else printf("\tfailed\n"); } catch (...) { if(numbers == NULL) printf("Test passed.\n"); else printf("Test failed.\n"); } } int _tmain(int argc, _TCHAR* argv[]) { // 典型輸入，單調升序的數組的一個旋轉
    int array1[] = {3, 4, 5, 1, 2}; Test(array1, sizeof(array1) / sizeof(int), 1); // 有重復數字，並且重復的數字剛好的最小的數字
    int array2[] = {3, 4, 5, 1, 1, 2}; Test(array2, sizeof(array2) / sizeof(int), 1); // 有重復數字，但重復的數字不是第一個數字和最后一個數字
    int array3[] = {3, 4, 5, 1, 2, 2}; Test(array3, sizeof(array3) / sizeof(int), 1); // 有重復的數字，並且重復的數字剛好是第一個數字和最后一個數字
    int array4[] = {1, 0, 1, 1, 1}; Test(array4, sizeof(array4) / sizeof(int), 0); // 單調升序數組，旋轉0個元素，也就是單調升序數組本身
    int array5[] = {1, 2, 3, 4, 5}; Test(array5, sizeof(array5) / sizeof(int), 1); // 數組中只有一個數字
    int array6[] = {2}; Test(array6, sizeof(array6) / sizeof(int), 2); // 輸入NULL
    Test(NULL, 0, 0); return 0; }

7. 參考代碼下載

項目 08_MinNumberInRotatedArray 下載： 百度網盤

何海濤《劍指Offer：名企面試官精講典型編程題》 所有參考代碼下載：百度網盤

 

參考資料

[1]  何海濤. 劍指 Offer：名企面試官精講典型編程題 [M]. 北京：電子工業出版社，2012. 63-71.