算法-搜索算法：二分查找（Binary Search）【前置条件：待查数据集必须是有序结构，可以右重复元素】【时间复杂度：O(logn)】

本文主要是介绍算法-搜索算法：二分查找（Binary Search）【前置条件：待查数据集必须是有序结构，可以右重复元素】【时间复杂度：O(logn)】，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

搜索：是在一个项目集合中找到一个特定项目的算法过程。搜索通常的答案是真的或假的，因为该项目是否存在。 搜索的几种常见方法：顺序/线性查找、二分法查找、二叉树查找、哈希查找

二分查找又称折半查找，优点是比较次数少，查找速度快，平均性能好；缺点是要求待查表：

• 必须采用顺序存储结构；
• 必须按关键字大小有序排列；
• 插入删除困难；

二分查找/折半查找方法适用于不经常变动而查找频繁的有序列表：

• 首先，假设表中元素是按升序排列，将表中间位置记录的关键字与查找关键字比较，如果两者相等，则查找成功；
• 否则利用中间位置记录将表分成前、后两个子表，如果中间位置记录的关键字大于查找关键字，则进一步查找前一子表，否则进一步查找后一子表。
• 重复以上过程，直到找到满足条件的记录，使查找成功，或直到子表不存在为止，此时查找不成功。

二分查找/折半查找的时间复杂度：O(logn)【不考虑排序的时间】

• 最差时间复杂度: O(logn)；
• 最优时间复杂度: O(1)
  

很多时候，和二分查找法相关的算法问题，不是在数组中查找特定值，而是使用二分查找法搜索问题的解。

一、二分法查找指定数值

1、递归方法

java版本：

public class BinarySearch {/*** 二分查找的时间复杂度是O(log(n))* 1.二分查找依赖顺序表结构即数组* 2.二分查找针对的是有序数据* 3.递归方法*/public static int search(int[] arr, int target) {return search(arr, 0, arr.length - 1, target);}private static int search(int[] arr, int left, int right, int target) {if (left > right)return -1;int mid = left + (right - left) / 2;System.out.println("target = " + target + "; arr[mid] = " + arr[mid] + "; mid = " + mid);if (arr[mid] == target) {return mid;}if (target < arr[mid]) {return search(arr, left, mid - 1, target);} else {return search(arr, mid + 1, right, target);}}public static void main(String[] args) {int arr[] = {3, 5, 11, 17, 21, 23, 101};int result = search(arr, 23);System.out.println("result = " + result);}
}

输出结果：

target = 23; arr[mid] = 17; mid = 3
target = 23; arr[mid] = 23; mid = 5
result = 5

python版本：

def binary_search(alist, item):"""二分查找,递归"""n = len(alist)if n > 0:mid = n // 2if alist[mid] == item:return Trueelif item < alist[mid]:return binary_search(alist[:mid], item)else:return binary_search(alist[mid + 1:], item)return Falseif __name__ == "__main__":li = [17, 20, 26, 31, 44, 54, 55, 77, 93]print(binary_search(li, 55))print(binary_search(li, 100))

2、非递归方法

C++版本：

#include <iostream>int binary_search(int arr[], int left, int right, int target) {while (left <= right) {int mid = left + (right - left) / 2;if (arr[mid] == target) {return mid;} else if (arr[mid] > target) {right = mid - 1;} else {left = mid + 1;}}
}int main() {int arr[] = {3, 5, 11, 17, 21, 23, 101};int len = sizeof(arr) / sizeof(arr[0]);int target = 23;int result = binary_search(arr, 0, len - 1, target);std::cout << "result = " << result << std::endl;
}

java版本：

public class BinarySearch02 {/*** 二分查找的时间复杂度是O(log(n))* 1.二分查找依赖顺序表结构即数组* 2.二分查找针对的是有序数据* 3.非递归方法*/public static int search(int[] arr, int target) {int left = 0;int right = arr.length - 1;// 循环不变量：在arr[left, right]的范围中查找 targetwhile (left < right) {int mid = left + (right - left) / 2;System.out.println("target = " + target + "; arr[mid] = " + arr[mid] + "; mid = " + mid);if (arr[mid] == target) {return mid;}// 如果没找到目标值，则修改左右边界if (arr[mid] < target) {left = mid + 1;} else {right = mid - 1;}}return -1;}public static void main(String[] args) {int arr[] = {3, 5, 11, 17, 21, 23, 101};int result = search(arr, 23);System.out.println("result = " + result);}
}

输出结果：

target = 23; arr[mid] = 17; mid = 3
target = 23; arr[mid] = 23; mid = 5
result = 5Process finished with exit code 0

python代码：方法一【循环不变量：在 $nums[left,right]$ 左闭右闭 区间查找target；初始范围是：$nums[0,len(nums)-1]$】

class Solution:"""二分查找， 非递归"""def upper(self, nums, target):# 初始化查找范围left, right = 0, len(nums) - 1# 循环不变量：在nums[left, right]左闭右闭区间查找target# 如果left<=right，说明要搜索的数组中还存在要搜索的元素，该元素有可能是我们要找的元素，需继续搜索while left <= right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left) // 2if nums[mid] == target:return midelif nums[mid] > target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找targetelse:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找targetprint("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return -1solution = Solution()
nums = [17, 20, 26, 31, 44, 54, 55, 77, 93]
target = 54
result = solution.upper(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 5---right = 8left = 5---right = 8最终：result =  6

python代码：方法二【循环不变量：在 $nums[left,right)$ 左闭右开 区间查找target；初始范围是：$nums[0,len(nums))$】

class Solution:"""二分查找， 非递归"""def upper(self, nums, target):# 初始化查找范围left, right = 0, len(nums)# 循环不变量：在nums[left, right)左闭右开区间查找target# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left) // 2if nums[mid] == target:return midelif nums[mid] > target:right = mid  # 继续在nums[left, mid) 范围内寻找targetelse:left = mid + 1  # 继续在nums[mid + 1, right) 范围内寻找targetprint("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return -1solution = Solution()
nums = [17, 20, 26, 31, 44, 54, 55, 77, 93]
target = 54
result = solution.upper(nums, target)
print("\n最终：result = ", result)

打印：

left = 0---right = 9
本轮循环结束：left = 5---right = 9left = 5---right = 9
本轮循环结束：left = 5---right = 7left = 5---right = 7
本轮循环结束：left = 5---right = 6left = 5---right = 6最终：result =  5

二、二分法查找法变种

1、upper：查找大于target的最小值

【循环不变量：在 $nums[left,right]$ 左闭右闭 区间查找target；初始范围是：$nums[0,len(nums)]$】

1.1 C++版本

#include <iostream>// 二分查找法：查找大于target的最小值int upper(int arr[], int left, int right, int target) {while (left < right) {int mid = left + (right - left) / 2;if (arr[mid] == target) {left = mid + 1;} else if (arr[mid] < target) {left = mid + 1;} else {right = mid;}}return left;
}int main() {int arr[] = {3, 5, 11, 17, 21, 23, 101, 101};int len = sizeof(arr) / sizeof(arr[0]);int target = 101;int result = upper(arr, 0, len, target);std::cout << "result = " << result << std::endl;
}

1.2 python版本：

class Solution:"""二分查找法：查找大于target的最小值"""def upper(self, nums, target):# 初始化查找范围left, right = 0, len(nums)# 循环不变量：在nums[left, right]左闭右开区间查找大于target的最小值# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left) // 2if nums[mid] == target:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找目标值【由于nums[mid] == target，所以mid处的元素绝不可能是要找的值，排除掉】elif nums[mid] < target:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找目标值【由于nums[mid] < target，所以mid处的元素绝不可能是要找的值，排除掉】else:   # nums[mid] > targetright = mid  # 继续在nums[left, mid] 范围内寻找目标值【此mid处的元素也有可能是答案，不能排除】print("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return left

比如：查找大于60的最小值

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 60
result = solution.upper(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 0---right = 4left = 0---right = 4
本轮循环结束：left = 3---right = 4left = 3---right = 4
本轮循环结束：left = 3---right = 3最终：result =  3

比如：查找大于99的最小值

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 99
result = solution.upper(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 5---right = 8left = 5---right = 8
本轮循环结束：left = 7---right = 8left = 7---right = 8
本轮循环结束：left = 8---right = 8最终：result =  8

比如：查找大于100的最小值

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 100
result = solution.upper(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 5---right = 8left = 5---right = 8
本轮循环结束：left = 7---right = 8left = 7---right = 8
本轮循环结束：left = 8---right = 8最终：result =  8

1.3 java版本：

public class BinarySearch_upper {/*** 查找 > target 的最小值的索引*/public static int upper(int[] arr, int target) {int left = 0;int right = arr.length;// 循环不变量：在arr[left, right]的范围中查找解while (left < right) {int mid = left + (right - left) / 2;System.out.println("target = " + target + "; arr[mid] = " + arr[mid] + "; mid = " + mid);if (arr[mid] <= target) {left = mid + 1;} else {right = mid;}}return left;}public static void main(String[] args) {int arr[] = {3, 5, 11, 17, 17, 21, 23, 101};int result = upper(arr, 11);System.out.println("result = " + result);}
}

输出结果：

target = 11; arr[mid] = 17; mid = 4
target = 11; arr[mid] = 11; mid = 2
target = 11; arr[mid] = 17; mid = 3
result = 3Process finished with exit code 0

2、lower：查找小于target的最大值

2.1 C++版本

#include <iostream>// 二分查找法：查找小于target的最大值int lower(int arr[], int left, int right, int target) {while (left < right) {int mid = left + (right - left + 1) / 2;if (arr[mid] == target) {right = mid - 1;} else if (arr[mid] > target) {right = mid - 1;} else {left = mid;}}return left;
}int main() {int arr[] = {3, 5, 11, 17, 21, 23, 101, 101};int len = sizeof(arr) / sizeof(arr[0]);int target = 17;int result = lower(arr, -1, len - 1, target);std::cout << "result = " << result << std::endl;
}

2.2 python版本【坑：$mid=left+(right-left+1)//2$下取整改为上取整，防止死循环】

class Solution:"""二分查找法：查找小于target的最大值"""def lower(self, nums, target):# 初始化查找范围left, right = -1, len(nums) - 1# 循环不变量：在nums[left, right]左闭右开区间查找小于target的最大值# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left + 1) // 2    # +1 防止 进入 left = mid 时下取整后进入死循环if nums[mid] == target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找目标值【由于nums[mid] == target，所以mid处的元素绝不可能是要找的值，排除掉】elif nums[mid] > target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找目标值【由于nums[mid] > target，所以mid处的元素绝不可能是要找的值，排除掉】else:  # nums[mid] < targetleft = mid  # 继续在nums[mid, right] 范围内寻找目标值【此mid处的元素也有可能是答案，不能排除】print("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return right

比如：查找小于60的最大值

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 60
result = solution.lower(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = -1---right = 2left = -1---right = 2
本轮循环结束：left = 1---right = 2left = 1---right = 2
本轮循环结束：left = 2---right = 2最终：result =  2

比如：查找小于23的最大值

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 23
result = solution.lower(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = -1---right = 2left = -1---right = 2
本轮循环结束：left = -1---right = 0left = -1---right = 0
本轮循环结束：left = -1---right = -1最终：result =  -1

比如：查找小于10的最大值

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 10
result = solution.lower(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = -1---right = 2left = -1---right = 2
本轮循环结束：left = -1---right = 0left = -1---right = 0
本轮循环结束：left = -1---right = -1最终：result =  -1

2.3 java版本

public class BinarySearch_lower {/*** 查找 < target 的最大值的索引*/public static int lower(int[] arr, int target) {int left = -1;int right = arr.length - 1;// 循环不变量：在arr[left, right]的范围中查找解while (left < right) {int mid = left + (right - left + 1) / 2;System.out.println("target = " + target + "; arr[mid] = " + arr[mid] + "; mid = " + mid);if (arr[mid] < target) {left = mid;} else {right = mid - 1;}}return left;}public static void main(String[] args) {int arr[] = {3, 5, 11, 17, 17, 21, 23, 101};int result = lower(arr, 11);System.out.println("result = " + result);}
}

输出结果：

target = 11; arr[mid] = 17; mid = 3
target = 11; arr[mid] = 5; mid = 1
target = 11; arr[mid] = 11; mid = 2
result = 1Process finished with exit code 0

三、二分法查找法变种组合：upper_ceil、lower_ceil

1、upper_ceil

• 如果存在元素，返回该元素最大索引

• 如果元素不存在，返回比该元素大的最小值

1.1 python版本

class Solution:"""二分查找法：查找大于target的最小值"""def upper(self, nums, target):# 初始化查找范围left, right = 0, len(nums)# 循环不变量：在nums[left, right]左闭右开区间查找大于target的最小值# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left) // 2if nums[mid] == target:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找目标值【由于nums[mid] == target，所以mid处的元素绝不可能是要找的值，排除掉】elif nums[mid] < target:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找目标值【由于nums[mid] < target，所以mid处的元素绝不可能是要找的值，排除掉】else:   # nums[mid] > targetright = mid  # 继续在nums[left, mid] 范围内寻找目标值【此mid处的元素也有可能是答案，不能排除】print("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return leftdef upper_ceil(self, nums, target):upper_idx = self.upper(nums, target)if upper_idx - 1 >= 0 and nums[upper_idx - 1] == target:upper_idx -= 1return upper_idx

比如：查找56

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 56
result = solution.upper_ceil(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 0---right = 4left = 0---right = 4
本轮循环结束：left = 3---right = 4left = 3---right = 4
本轮循环结束：left = 3---right = 3最终：result =  2

比如：查找60

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 60
result = solution.upper_ceil(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 0---right = 4left = 0---right = 4
本轮循环结束：left = 3---right = 4left = 3---right = 4
本轮循环结束：left = 3---right = 3最终：result =  3

1.2 java版本：

public class BinarySearch_ceil {/*** 查找 > target 的最小值的索引*/public static int upper(int[] arr, int target) {int left = 0;int right = arr.length;// 循环不变量：在arr[left, right]的范围中查找解while (left < right) {int mid = left + (right - left) / 2;System.out.println("target = " + target + "; arr[mid] = " + arr[mid] + "; mid = " + mid);if (arr[mid] <= target) {left = mid + 1;} else {right = mid;}}return left;}public static int ceil(int[] arr, int target) {int u = upper(arr, target);if (u - 1 >= 0 && arr[u - 1] == target) {return u - 1;}return u;}public static void main(String[] args) {int arr[] = {3, 5, 11, 17, 17, 21, 23, 101};int result01 = ceil(arr, 11);System.out.println("result01 = " + result01);System.out.println("===========================================");int result02 = ceil(arr, 13);System.out.println("result02 = " + result02);}
}

输出结果：

target = 11; arr[mid] = 17; mid = 4
target = 11; arr[mid] = 11; mid = 2
target = 11; arr[mid] = 17; mid = 3
result01 = 2
===========================================
target = 13; arr[mid] = 17; mid = 4
target = 13; arr[mid] = 11; mid = 2
target = 13; arr[mid] = 17; mid = 3
result02 = 3Process finished with exit code 0

2、lower_ceil（>=target的最小索引）

• 如果该元素存在，返回该元素最小索引

• 如果该元素不存在，返回大于该元素的最小索引

class Solution:def upper(self, nums, target):# 初始化查找范围left, right = 0, len(nums)# 循环不变量：在nums[left, right]左闭右开区间查找大于target的最小值# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left) // 2if nums[mid] == target:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找目标值【由于nums[mid] == target，所以mid处的元素绝不可能是要找的值，排除掉】elif nums[mid] < target:left = mid + 1  # 继续在nums[mid + 1, right] 范围内寻找目标值【由于nums[mid] < target，所以mid处的元素绝不可能是要找的值，排除掉】else:  # nums[mid] > targetright = mid  # 继续在nums[left, mid] 范围内寻找目标值【此mid处的元素也有可能是答案，不能排除】print("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return leftdef lower_ceil(self, nums, target):upper_idx = self.upper(nums, target)while upper_idx - 1 >= 0 and nums[upper_idx - 1] == target:upper_idx -= 1return upper_idx

比如：查找56

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 56
result = solution.lower_ceil(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 0---right = 4left = 0---right = 4
本轮循环结束：left = 3---right = 4left = 3---right = 4
本轮循环结束：left = 3---right = 3最终：result =  1

比如：查找60

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 60
result = solution.lower_ceil(nums, target)
print("\n最终：result = ", result)

打印结果：

left = 0---right = 8
本轮循环结束：left = 0---right = 4left = 0---right = 4
本轮循环结束：left = 3---right = 4left = 3---right = 4
本轮循环结束：left = 3---right = 3最终：result =  3

四、二分法查找法变种组合：lower_floor、upper_floor

1、lower_floor

• 如果该元素存在，返回该元素最小索引

• 如果该元素不存在，返回小于该元素的最大索引

class Solution:def lower(self, nums, target):# 初始化查找范围left, right = -1, len(nums) - 1# 循环不变量：在nums[left, right]左闭右开区间查找小于target的最大值# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left + 1) // 2    # +1 防止 进入 left = mid 时下取整后进入死循环if nums[mid] == target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找目标值【由于nums[mid] == target，所以mid处的元素绝不可能是要找的值，排除掉】elif nums[mid] > target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找目标值【由于nums[mid] > target，所以mid处的元素绝不可能是要找的值，排除掉】else:  # nums[mid] < targetleft = mid  # 继续在nums[mid, right] 范围内寻找目标值【此mid处的元素也有可能是答案，不能排除】print("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return rightdef lower_floor(self, nums, target):lower_idx = self.lower(nums, target)if lower_idx + 1 <= len(nums) - 1 and nums[lower_idx + 1] == target:lower_idx += 1return lower_idx

比如：查找69

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 69
result = solution.lower_floor(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = 3---right = 7left = 3---right = 7
本轮循环结束：left = 3---right = 4left = 3---right = 4
本轮循环结束：left = 3---right = 3最终：result =  4

比如：查找60

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 60
result = solution.lower_floor(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = -1---right = 2left = -1---right = 2
本轮循环结束：left = 1---right = 2left = 1---right = 2
本轮循环结束：left = 2---right = 2最终：result =  2

2、upper_floor（<=target的最大索引）

• 如果该元素存在，返回该元素最大索引

• 如果该元素不存在，返回小于该元素的最大索引

class Solution:def lower(self, nums, target):# 初始化查找范围left, right = -1, len(nums) - 1# 循环不变量：在nums[left, right]左闭右开区间查找小于target的最大值# 由于待搜索数组空间为左闭右开，如果当left == right，说明要搜索的数组已经是空数组while left < right:print("left = {0}---right = {1}".format(left, right))mid = left + (right - left + 1) // 2    # +1 防止 进入 left = mid 时下取整后进入死循环if nums[mid] == target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找目标值【由于nums[mid] == target，所以mid处的元素绝不可能是要找的值，排除掉】elif nums[mid] > target:right = mid - 1  # 继续在nums[left, mid - 1] 范围内寻找目标值【由于nums[mid] > target，所以mid处的元素绝不可能是要找的值，排除掉】else:  # nums[mid] < targetleft = mid  # 继续在nums[mid, right] 范围内寻找目标值【此mid处的元素也有可能是答案，不能排除】print("本轮循环结束：left = {0}---right = {1}\n".format(left, right))return rightdef upper_floor(self, nums, target):lower_idx = self.lower(nums, target)while lower_idx + 1 <= len(nums) - 1 and nums[lower_idx + 1] == target:lower_idx += 1return lower_idx

比如：查找60

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 60
result = solution.upper_floor(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = -1---right = 2left = -1---right = 2
本轮循环结束：left = 1---right = 2left = 1---right = 2
本轮循环结束：left = 2---right = 2最终：result =  2

比如：查找56

solution = Solution()
nums = [23, 56, 56, 65, 69, 69, 99, 99]
target = 56
result = solution.upper_floor(nums, target)
print("\n最终：result = ", result)

打印结果：

left = -1---right = 7
本轮循环结束：left = -1---right = 2left = -1---right = 2
本轮循环结束：left = -1---right = 0left = -1---right = 0
本轮循环结束：left = 0---right = 0最终：result =  2

五、二分法查找法模板

这篇关于算法-搜索算法：二分查找（Binary Search）【前置条件：待查数据集必须是有序结构，可以右重复元素】【时间复杂度：O(logn)】的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

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