Description

Given a sorted array of n integers, find the starting and ending position of a given target value.

If the target is not found in the array, return [-1, -1].

Example

Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Challenge

O(log n) time.

这个题目的思路就是用两个binary search, 分别求first index 和last index.

Code   其实可以利用helper funciton来去优化code.

class Solution:
def searchRange(self, A, target):
# write your code here
if not A: return [-1,-1]
l, r , ans = 0, len(A) -1, [-1,-1]
while l + 1 < r:
mid = l + (r - l)//2
if A[mid] < target:
l = mid
elif A[mid] > target:
r = mid
else:
r = mid
if A[l] == target:
ans[0] = l
elif A[r] == target:
ans[0] = r
else:
return ans # find last index
l, r = 0, len(A) -1
while l + 1 < r:
mid = l + (r - l)//2
if A[mid] < target:
l = mid
elif A[mid] > target:
r = mid
else:
l = mid
if A[r] == target:
ans[1] = r
elif A[l] == target:
ans[1] = l
else:
return ans
return ans

去掉不必要的行

class Solution:
def searchRange(self, A, target):
# write your code here
if not A: return [-1,-1]
l, r , ans = 0, len(A) -1, [-1,-1]
while l + 1 < r:
mid = l + (r - l)//2
if A[mid] < target:
l = mid
else:
r = mid
if A[l] == target: ans[0] = l
elif A[r] == target: ans[0] = r
else:
return ans # find last index
l, r = 0, len(A) -1
while l + 1 < r:
mid = l + (r - l)//2
if A[mid] <= target:
l = mid
else:
r = mid
if A[r] == target: ans[1] = r
elif A[l] == target: ans[1] = l
return ans

Use helper function to make the code even shorter.

class Solution:
def searchRange(self, A, target):
l, r = 0, len(A) - 1
if not A or nums[l] > target or nums[r] < target:
return [-1, -1]
def helper(points, pos, target):
while points[0] + 1 < points[1]:
mid = points[0] + (points[1] - points[0])//2
if A[mid] > target:
points[1] = mid
elif A[mid] < target:
points[0] = mid
else:
points[pos] = mid
if A[points[1 - pos]] == target: return points[1 - pos]
if A[points[pos]] == target: return points[pos]
return -1
return [helper([l, r], 1, target), helper([l, r], 0, target)]

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