题解 | #二分查找-II#
二分查找-II
https://www.nowcoder.com/practice/4f470d1d3b734f8aaf2afb014185b395
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
# 如果目标值存在返回下标,否则返回 -1
# @param nums int整型一维数组
# @param target int整型
# @return int整型
#
class Solution:
def search(self , nums: List[int], target: int) -> int:
if len(nums) == 0:
return -1
low = 0
high = len(nums) - 1
# 全局一致
if nums[0] == target:
return 0
while low <= high:
mid = low + (high - low) // 2
if nums[mid] > target:
high = mid - 1
elif nums[mid] < target:
low = mid + 1
else:
if mid==0 or nums[mid-1] < target:
return mid
else:
high = mid - 1
return -1
注意点:
1、空数组
2、最后的else需要注意mid-1不要溢出,增加mid==0判断