给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。
如果数组中不存在目标值 target，返回 [-1, -1]。
进阶：
你可以设计并实现时间复杂度为 O(log n) 的算法解决此问题吗？
示例 1：
输入：nums = [5,7,7,8,8,10], target = 8
输出：[3,4]
示例 2：
输入：nums = [5,7,7,8,8,10], target = 6
输出：[-1,-1]
示例 3：
输入：nums = [], target = 0
输出：[-1,-1]
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
python
class Solution:
def searchRange(self, nums: [int], target: int) -> [int]:
# 二分法，时间O(logn), 空间O(1)
def search(nums: [int], target: int) -> int:
left, right = 0, len(nums)
while left < right:
mid = int((left+right)/2)
if nums[mid] >= target:
right = mid
else:
left = mid + 1
return left
# 找到第一个=target的位置
leftIdx = search(nums, target)
# 找到第一个 > target的位置
rightIdx = search(nums, target+1)
if leftIdx == len(nums) or nums[leftIdx] != target:
return [-1, -1]
return [leftIdx, rightIdx-1]
def searchRange1(self, nums, target):
"""
暴力遍历法， 时间O(n)
:param nums:
:param target:
:return:
"""
record = list()
for i in range(len(nums)):
if nums[i] == target:
record.append(i)
if len(record) > 1:
firstIdx = record[0]
lastIdx = record[-1]
else:
return [-1, -1]
return [firstIdx, lastIdx]
if __name__ == "__main__":
nums1 = [1,3]
target1 = 2
nums2 = [1,3,3,3,3,3,3,3,5,5,6]
target2 = 5
test = Solution()
print(test.searchRange(nums1,target1)) # [-1, -1]
print(test.searchRange(nums2,target2)) # [8, 9]
print(test.searchRange1(nums2,target2)) # [8, 9]
print(test.searchRange1(nums1,target1)) # [-1, -1]
golang
package main
import "fmt"
func main() {
var nums = []int{1, 3, 3, 3, 3, 3, 3, 3, 5, 5, 6}
var target int = 5
res := searchRange(nums, target)
fmt.Println(res)
res1 := searchRange1(nums, target)
fmt.Println(res1)
}
// 二分法，时间O(logn), 空间O(1)
func searchRange(nums []int, target int) []int {
//第一个=target的位置
leftIdx := search(nums, target)
rightIdx := search(nums, target+1)
if leftIdx == len(nums) || nums[leftIdx] != target {
return []int{-1, -1}
}
return []int{leftIdx, rightIdx - 1}
}
func search(nums []int, target int) int {
var left int = 0
var right int = len(nums)
for left < right {
var mid int = (left + right) / 2
if nums[mid] >= target {
right = mid
} else {
left = mid + 1
}
}
return left
}
// 暴力法， 时间O(n)，空间有限变量
func searchRange1(nums []int, target int) []int {
var firstIdx int
var lastIdx int
var record = []int{}
for i, _ := range nums {
if nums[i] == target {
record = append(record, i)
}
}
if len(record) > 1 {
firstIdx = record[0]
lastIdx = record[len(record)-1]
} else {
return []int{-1, -1}
}
return []int{firstIdx, lastIdx}
}