LeetCode #34 Search for a Range
LeetCode #34 Search for a Range
Question
Given an array of integers sorted in ascending order, find the starting and ending position of a given target value.
Your algorithm's runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1]
.
For example,
Given [5, 7, 7, 8, 8, 10]
and target value 8,
return [3, 4]
.
Solution
Approach #1
class Solution {
func searchRange(_ nums: [Int], _ target: Int) -> [Int] {
let index = searchNumber(nums, target, 0, nums.count - 1)
if index < 0 {
return [-1, -1]
}
var l = searchNumber(nums, target, 0, index)
while l >= 0 {
let t = searchNumber(nums, target, 0, l - 1)
if t < 0 {
break
}
l = t
}
var h = searchNumber(nums, target, index, nums.count - 1)
while h >= 0 {
let t = searchNumber(nums, target, h + 1, nums.count - 1)
if t < 0 {
break
}
h = t
}
return [l, h]
}
func searchNumber(_ nums: [Int], _ target: Int, _ low: Int, _ high: Int) -> Int {
var l = low
var h = high
var m = 0
while l <= h {
m = l + (h - l) / 2
if target < nums[m] {
h = m - 1
} else if target > nums[m] {
l = m + 1
} else {
return m
}
}
return -1
}
}
Time complexity: O(log(n)).
Space complexity: O(1).
Approach #2
class Solution {
func searchRange(_ nums: [Int], _ target: Int) -> [Int] {
var result: [Int] = [-1, -1]
if nums.isEmpty { return result }
var l = 0
var h = nums.count - 1
while l < h {
let m = l + (h - l) / 2
if target > nums[m] {
l = m + 1
} else if target < nums[m] {
h = m - 1
} else {
h = m
}
}
if nums[l] != target { return result }
result[0] = l
h = nums.count - 1
while l < h {
let m = l + (h - l) / 2 + 1
if target < nums[m] {
h = m - 1
} else if target > nums[m] {
l = m + 1
} else {
l = m
}
}
result[1] = l
return result
}
}
Time complexity: O(log(n)).
Space complexity: O(1).
转载请注明出处:http://www.cnblogs.com/silence-cnblogs/p/7067307.html