题目描述
给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。
如果数组中不存在目标值 target，返回 [-1, -1]。
你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。
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]
示例 3：
输入：nums = [], target = 0
输出：[-1,-1]
提示：
0 <= nums.length <= 105-109 <= nums[i] <= 109nums 是一个非递减数组-109 <= target <= 109
解法
方法一：二分查找
我们可以进行两次二分查找，分别查找出左边界和右边界。
时间复杂度 $O(\log n)$，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$。
| class Solution:
def searchRange(self, nums: List[int], target: int) -> List[int]:
l = bisect_left(nums, target)
r = bisect_left(nums, target + 1)
return [-1, -1] if l == r else [l, r - 1]
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20 | class Solution {
public int[] searchRange(int[] nums, int target) {
int l = search(nums, target);
int r = search(nums, target + 1);
return l == r ? new int[] {-1, -1} : new int[] {l, r - 1};
}
private int search(int[] nums, int x) {
int left = 0, right = nums.length;
while (left < right) {
int mid = (left + right) >>> 1;
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}
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| class Solution {
public:
vector<int> searchRange(vector<int>& nums, int target) {
int l = lower_bound(nums.begin(), nums.end(), target) - nums.begin();
int r = lower_bound(nums.begin(), nums.end(), target + 1) - nums.begin();
if (l == r) {
return {-1, -1};
}
return {l, r - 1};
}
};
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| func searchRange(nums []int, target int) []int {
l := sort.SearchInts(nums, target)
r := sort.SearchInts(nums, target+1)
if l == r {
return []int{-1, -1}
}
return []int{l, r - 1}
}
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17 | function searchRange(nums: number[], target: number): number[] {
const search = (x: number): number => {
let [left, right] = [0, nums.length];
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};
const l = search(target);
const r = search(target + 1);
return l === r ? [-1, -1] : [l, r - 1];
}
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24 | impl Solution {
pub fn search_range(nums: Vec<i32>, target: i32) -> Vec<i32> {
let n = nums.len();
let search = |x| {
let mut left = 0;
let mut right = n;
while left < right {
let mid = left + (right - left) / 2;
if nums[mid] < x {
left = mid + 1;
} else {
right = mid;
}
}
left
};
let l = search(target);
let r = search(target + 1);
if l == r {
return vec![-1, -1];
}
vec![l as i32, (r - 1) as i32]
}
}
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23 | /**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var searchRange = function (nums, target) {
function search(x) {
let left = 0,
right = nums.length;
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] >= x) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
const l = search(target);
const r = search(target + 1);
return l == r ? [-1, -1] : [l, r - 1];
};
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26 | class Solution {
/**
* @param Integer[] $nums
* @param Integer $target
* @return Integer[]
*/
function searchRange($nums, $target) {
$search = function ($x) use ($nums) {
$left = 0;
$right = count($nums);
while ($left < $right) {
$mid = intdiv($left + $right, 2);
if ($nums[$mid] >= $x) {
$right = $mid;
} else {
$left = $mid + 1;
}
}
return $left;
};
$l = $search($target);
$r = $search($target + 1);
return $l === $r ? [-1, -1] : [$l, $r - 1];
}
}
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21 | class Solution {
fun searchRange(nums: IntArray, target: Int): IntArray {
val left = this.search(nums, target)
val right = this.search(nums, target + 1)
return if (left == right) intArrayOf(-1, -1) else intArrayOf(left, right - 1)
}
private fun search(nums: IntArray, target: Int): Int {
var left = 0
var right = nums.size
while (left < right) {
val middle = (left + right) / 2
if (nums[middle] < target) {
left = middle + 1
} else {
right = middle
}
}
return left
}
}
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