1、问题描述
给定一个排序的整数数组(升序)和一个要查找的目标整数target,查找到第1次出现的下标(从0开始),如果target不存在于数组中,返回-1。
2、问题示例
输入数组[1,4,4,5,7,7,8,9,9,10],和目标整数1,输出其所在的位置为0,即第一次出现在第0个位置。
输入数组[1,2,3,3,4,5,10],和目标整数3,输出其所在的位置为2,即第一次出现在第2个位置。
输入数组[1,2,3,3,4,5,10],和目标整数6,输出其所在的位置为-1,即没有出现过6,返回-1。
3、代码实现
class Solution:
#参数nums:整数数组
#参数target:要查找的目标数字
#返回值:目标数字的第一个位置,从0开始
def binarySearch(self, nums, target):
return self.search(nums, 0, len(nums) - 1, target)
def search(self, nums, start, end, target):
if start > end:
return - 1
mid = (start + end)//2
if nums[mid] > target:
return self.search(nums, start, mid, target)
if nums[mid] == target:
return mid
if nums[mid] < target:
return self.search(nums, mid, end, target)
if __name__ == '__main__':
my_solution = Solution()
nums = [1,4,4,5,7,7,8,9,9,10]
target = 1
targetIndex = my_solution.binarySearch(nums, target)
print("输入:nums = ", nums ," " , "target =", target)
print("输出: ",targetIndex)
4、上述代码是书本例子,但发现如果target不在数组中出现,会产生RecursionError: maximum recursion depth exceeded in comparison的错误,以下为修改后的代码:
class Solution:
def binarySearch(self, nums, start, end, target):
start = 0
end = len(nums)
if start > end:
mid = (start + end)//2
if nums[mid] > target:
return self.binarySearch(nums, start, mid-1, target)
if nums[mid] == target:
return mid
if nums[mid] < target:
return self.binarySearch(nums, mid+1, end, target)
return -1
if __name__ == '__main__':
my_solution = Solution()
nums = [1,4,4,5,7,7,8,9,9,10]
target = 3
targetIndex = my_solution.binarySearch(nums, 0, len(nums), target)
print("输入:nums = ", nums ," " , "target =", target)
print("输出: ",targetIndex)