C++ 和 Python 达成旋转数组的最小数字
发布时间：2021-11-12 13:34:29 所属栏目：教程 来源：互联网
导读：把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如数组 {3, 4, 5, 1, 2} 为 {1, 2, 3, 4, 5} 的一个旋转，该数组的最小值为 1 。 算法设计思想 1. 暴力查找（Bruteforce
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把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个递增排序的数组的一个旋转，输出旋转数组的最小元素。例如数组 {3, 4, 5, 1, 2} 为 {1, 2, 3, 4, 5} 的一个旋转，该数组的最小值为 1 。 算法设计思想 1. 暴力查找（Bruteforce Search）：把旋转数组从前到后遍历一遍，其时间复杂度为 O(n)。很明显，这种思想非常直接粗暴，没有用到旋转数组的性质。 2. 二分查找（Binary Search）：每次查找都把旋转数组平均分成两部分，通过比较当前旋转数组两端点和中间点的值，判断最小值在数组的哪一部分，从而达到缩小搜索范围的目的。其中，两端点为当前的旋转数组索引最小和索引最大的元素，中间点为这两部分子数组的连结元素，也可以看做为轴枢点（pivot point），这里取旋转数组的最小索引和最大索引的算术平均值（向下取整）作为索引，其对应的元素作为中间点。具体过程，如图 2.10 所示。 需要注意，当旋转数组的两端点的值都与中间点的值相等时，因为无法判断最小值在哪一部分，因此需要采用顺序查找方法，即暴力查找。其示例如图 2.11 所示。 C++ 实现 #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) // 易错点 { // 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; } C++ 和 Python 实现旋转数组的最小数字 Python 实现 #!/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() 参考代码 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. // // 《剑指Offer——名企面试官精讲典型编程题》代码 // 著作权所有者：何海涛 #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指向相邻的两个数， // 则index1指向第一个递增子数组的最后一个数字， // index2指向第二个子数组的第一个数字，也就是数组中的最小数字 if(index2 - index1 == 1) { indexMid = index2; break; } // 如果下标为index1、index2和indexMid指向的三个数字相等， // 则只能顺序查找 indexMid = (index1 + index2) / 2; if(numbers[index1] == numbers[index2] && numbers[indexMid] == numbers[index1]) return MinInOrder(numbers, index1, index2); // 缩小查找范围 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; } // ====================测试代码==================== 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("tpassedn"); else printf("tfailedn"); } catch (...) { if(numbers == NULL) printf("Test passed.n"); else printf("Test failed.n"); } } int _tmain(int argc, _TCHAR* argv[]) { // 典型输入，单调升序的数组的一个旋转 int array1[] = {3, 4, 5, 1, 2}; Test(array1, sizeof(array1) / sizeof(int), 1); // 有重复数字，并且重复的数字刚好的最小的数字 int array2[] = {3, 4, 5, 1, 1, 2}; Test(array2, sizeof(array2) / sizeof(int), 1); // 有重复数字，但重复的数字不是第一个数字和最后一个数字 int array3[] = {3, 4, 5, 1, 2, 2}; Test(array3, sizeof(array3) / sizeof(int), 1); // 有重复的数字，并且重复的数字刚好是第一个数字和最后一个数字 int array4[] = {1, 0, 1, 1, 1}; Test(array4, sizeof(array4) / sizeof(int), 0); // 单调升序数组，旋转0个元素，也就是单调升序数组本身 int array5[] = {1, 2, 3, 4, 5}; Test(array5, sizeof(array5) / sizeof(int), 1); // 数组中只有一个数字 int array6[] = {2}; Test(array6, sizeof(array6) / sizeof(int), 2); // 输入NULL Test(NULL, 0, 0); return 0; }  （编辑：济南站长网） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |
