解法一：二分查找+左右线性探测
找到一个 target，然后向左或向右线性搜索左右边界
func searchRange(nums []int, target int) []int {
idx := binarySearch(nums, target)
if idx == -1{
return []int{-1, -1}
}
// 从该位置左右扩散，找到左右边界
leftbound := 0
for i := idx; i >=0; i--{
if nums[i] != target{ // 找到左边第一个不等于target的
leftbound = i+1
break
}
}
rightbound := len(nums)-1
for j := idx; j<len(nums); j++{
if nums[j] != target{ // 找到右边第一个不等于target的
rightbound = j-1
break
}
}
return []int{leftbound, rightbound}
}
// 二分查找目标值所处的位置，不存在返回-1
func binarySearch(nums []int, target int) int {
left, right := 0, len(nums)-1
for left <= right{
mid := left + (right - left)/2
if nums[mid] == target{
return mid
}else if nums[mid] < target{
left = mid+1
}else{
right=mid-1
}
}
return -1
}
但是这个解法不符合题目要求的时间复杂度
解法二：寻找左右边界
func searchRange(nums []int, target int) []int {
left := leftBound(nums, target)
right := rightBound(nums, target)
return []int{left, right}
}
func leftBound(nums []int, target int) int{
left, right := 0, len(nums)-1
for left <= right { // 退出条件 left - 1 = right
mid := left + (right-left)/2
if nums[mid] == target{ // 收缩右边界
right = mid -1
}else if nums[mid] < target {
left = mid + 1
}else if nums[mid] > target {
right = mid - 1
}
}
if left < 0 || left >= len(nums){ // 找不到target，循环退出条件left已越界
return -1
}
if nums[left] == target{
return left
}
return -1
}
func rightBound(nums []int, target int) int{
left, right := 0, len(nums)-1
for left <= right{ // 退出条件 left - 1 = right
mid := left + (right -left)/2
if nums[mid] == target{ // 收缩左边界
left = mid + 1
}else if nums[mid] < target{
left = mid + 1
}else if nums[mid] > target {
right = mid - 1
}
}
if right < 0 || right >= len(nums){ // 找不到target，循环退出条件right已越界
return -1
}
if nums[right] == target{
return right
}
return -1
}
