[LeetCode] 34. 在排序数组中查找元素的第一个和最后一个位置
题目链接 :https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/
题目描述:
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。
你的算法时间复杂度必须是 O(log n) 级别。
如果数组中不存在目标值，返回 [-1, -1]。
示例:
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]
思路:
思路1:手动二分法!
这里提供两种方法,一种容易理解,一种万能公式!
版本1:是指在二分法进行时,就判读是否有等于target的,但是找到的target不知道是最左边的数还是最右边的数,所以,通过找到这个数再找出相应的边界值.
版本2:是指二分法执行完毕,返回target在最左边的位置,在求出另一个边界!
关于详细说明,请看这篇[二分搜索](二分查找有几种写法？它们的区别是什么？ - Jason Li的回答 - 知乎
https://www.zhihu.com/question/36132386/answer/530313852),读完之后,可以加深二分搜索的理解!
思路2:库函数
python 的 bisect库
简要介绍一下,
bisect.bisect_left(a,x,lo=0,hi=len(a))在a中找x最左边数的索引,如果找不到就返回插入的索引.
``bisect.bisect(a, x, lo=0, hi=len(a))`找最右边的!
思路3:
当然上面的时间复杂度都是:\(O(logn)\)
关注我的知乎专栏,了解更多的解题技巧,大家共同进步!
代码:
思路1:
版本1python
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
if not nums:
return [-1, -1]
res = [-1, -1]
left = 0
n = len(nums)
right = n - 1
while left <= right:
mid = left + (right - left) // 2
if nums[mid] == target:
left = mid
right = mid
while left > 0 and nums[left-1] == nums[left]:
left -= 1
while right < n - 1 and nums[right] == nums[right+1]:
right += 1
res[0] = left
res[1] = right
break
elif nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return res
版本1java
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] res = {-1, -1};
int n = nums.length;
if (n == 0) return res;
int left = 0;
int right = n-1;
while (left <= right){
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
left = mid;
right = mid;
while (left > 0 && nums[left] == nums[left-1]) left--;
while (right < n - 1 && nums[right] == nums[right+1]) right++;
res[0] = left;
res[1] = right;
return res;
}
else if (nums[mid] > target) right = mid - 1;
else left = mid + 1;
}
return res;
}
}
思路1
版本2python
class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
if not nums:
return [-1, -1]
res = [-1, -1]
left = 0
n = len(nums)
right = n
while left < right:
mid = left + (right - left) // 2
if nums[mid] < target:
left = mid + 1
else:
right = mid
res[0] = left if left < n and nums[left] == target else -1
if res[0] == -1:
return res
while left < n - 1 and nums[left] == nums[left + 1]:
left += 1
return res
版本2java
class Solution {
public int[] searchRange(int[] nums, int target) {
int[] res = {-1, -1};
int n = nums.length;
if (n == 0) return res;
int left = 0;
int right = n;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] < target) left = mid + 1;
else right = mid;
}
res[0] = (left < n && nums[left] == target) ? left : -1;
if (res[0] == -1) return res;
while (left < n - 1 && nums[left] == nums[left + 1]) left++;
res[1] = left;
return res;
}
}
对库函数使用,大家自行操作的!
