python实现的lower_bound和upper_bound

时间:2022-03-02 13:06:12

1. lower_bound(nums, target)

在非递减数组nums中,lower_bound(nums, target)返回第一个大于等于target的值得位置,如果nums中元素均小于target(即不存在>=target的元素),则返回nums的长度(即target如果要插入到nums中,应该插入的位置)

#coding=utf-8
#返回nums中第一个>=target的值得位置,如果nums中都比target小,则返回len(nums)
def lower_bound(nums, target):
low, high = 0, len(nums)-1
pos = len(nums)
while low<high:
mid = (low+high)/2
if nums[mid] < target:
low = mid+1
else:#>=
high = mid
pos = high
return pos
测试

nums = [10, 10, 10, 20, 20, 20, 30, 30]   
print lower_bound(nums, 9),
print lower_bound(nums, 10),
print lower_bound(nums, 15),
print lower_bound(nums, 20),
print lower_bound(nums, 25),
print lower_bound(nums, 30),
print lower_bound(nums, 40)
运行结果:

python实现的lower_bound和upper_bound


2. upper_bound(nums, target)

在非递减数组nums中,upper_bound(nums, target)返回第一个大于target的值位置,如果nums中元素均小于等于target(即不存在>target的元素),则返回nums的长度(即target如果要插入到nums中,应该插入的位置)

#返回nums中第一个>target的值得位置,如果nums中都不比target大,则返回len(nums)
def upper_bound(nums, target):
low, high = 0, len(nums)-1
pos = len(nums)
while low<high:
mid=(low+high)/2
if nums[mid]<=target:
low = mid+1
else:#>
high = mid
pos = high
return pos
测试:

nums = [10, 10, 10, 20, 20, 20, 30, 30] 
print upper_bound(nums, 9),
print upper_bound(nums, 10),
print upper_bound(nums, 15),
print upper_bound(nums, 20),
print upper_bound(nums, 25),
print upper_bound(nums, 30),
print upper_bound(nums, 40)
运行结果:

python实现的lower_bound和upper_bound


c++ lower_bound源码参考:http://www.cplusplus.com/reference/algorithm/lower_bound/

c++ upper_bound源码参考:http://www.cplusplus.com/reference/algorithm/upper_bound/