lintcode： search for a range 搜索区间

题目

搜索区间

给定一个包含 n 个整数的排序数组，找出给定目标值 target 的起始和结束位置。

如果目标值不在数组中，则返回

[-1, -1]

样例

给出

[5, 7, 7, 8, 8, 10]

和目标值target=

8

,

返回

[3, 4]

挑战

时间复杂度 O(log n)

解题

正常解法是二分法分别找左右边界的，时间复杂度O(logN),但是要注意边界，边界很多种情况的。

找左边界：

1.起始位置是0的单独考虑：nums[0] == target return 0

2.正常的二分查找思想

　　　while(left <= right){
int median = (left + right)/2;
if(nums[median] < target){
left = median+1;
}else if(nums[median]>target){
right = median -1;
}else if(nums[median]==target){
if(nums[median-1]!=target)
return median;
else
right = median - 1;
}
}

找右边界：

1.结束位置单独考虑：nums[right] == target return right

2.正常的二分查找思想

while(left <= right){
int median = (left + right)/2;
if(nums[median] < target){
left = median+1;
}else if(nums[median]>target){
right = median -1;
}else if(nums[median]==target){
if(nums[median+1]!=target)
return median;
else
left = median + 1;
}
}

Java

class Solution:
"""
@param A : a list of integers
@param target : an integer to be searched
@return : a list of length 2, [index1, index2]
"""
def searchRange(self, A, target):
# write your code here
res = [-1,-1]
if A==None or len(A) == 0:
return res
res[0] = self.BoundaryLeft(A,0,len(A)-1,target)
res[1] = self.BoundaryRight(A,0,len(A)-1,target)
return res
def BoundaryLeft(self,A,left,right,target):
if A[0] == target:
return 0
while left<= right:
median = (left + right)/2
if A[median] < target:
left = median + 1
elif A[median] > target:
right = median -1
elif A[median] == target:
if A[median-1]!=target:
return median
else:
right = median -1
return -1

def BoundaryRight(self,A,left,right,target):
if A[right] == target:
return right
while left<= right:
median = (left+ right)/2
if A[median] < target:
left = median + 1
elif A[median] > target:
right = median - 1
elif A[median] == target:
if A[median + 1]!=target:
return median
else:
left = median + 1
return -1

Python Code