题目描述
给定一个排序的整数数组 nums
和一个整数目标值 target
,请在数组中找到 target
,并返回其下标。如果目标值不存在于数组中,返回它将会被按顺序插入的位置。
请必须使用时间复杂度为 O(log n)
的算法。
示例 1:
输入: nums = [1,3,5,6], target = 5
输出: 2
示例 2:
输入: nums = [1,3,5,6], target = 2
输出: 1
示例 3:
输入: nums = [1,3,5,6], target = 7
输出: 4
示例 4:
输入: nums = [1,3,5,6], target = 0
输出: 0
示例 5:
输入: nums = [1], target = 0
输出: 0
提示:
1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums
为无重复元素的升序排列数组
-104 <= target <= 104
注意:本题与主站 35 题相同: https://leetcode.cn/problems/search-insert-position/
解法
方法一
| class Solution:
def searchInsert(self, nums: List[int], target: int) -> int:
left, right = 0, len(nums)
while left < right:
mid = (left + right) >> 1
if nums[mid] >= target:
right = mid
else:
left = mid + 1
return left
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14 | class Solution {
public int searchInsert(int[] nums, int target) {
int left = 0, right = nums.length;
while (left < right) {
int mid = (left + right) >>> 1;
if (nums[mid] >= target) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14 | class Solution {
public:
int searchInsert(vector<int>& nums, int target) {
int left = 0, right = nums.size();
while (left < right) {
int mid = left + right >> 1;
if (nums[mid] >= target)
right = mid;
else
left = mid + 1;
}
return left;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12 | func searchInsert(nums []int, target int) int {
left, right := 0, len(nums)
for left < right {
mid := (left + right) >> 1
if nums[mid] >= target {
right = mid
} else {
left = mid + 1
}
}
return left
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18 | /**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var searchInsert = function (nums, target) {
let left = 0;
let right = nums.length;
while (left < right) {
const mid = (left + right) >> 1;
if (nums[mid] >= target) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15 | class Solution {
func searchInsert(_ nums: [Int], _ target: Int) -> Int {
var left = 0
var right = nums.count
while left < right {
let mid = (left + right) / 2
if nums[mid] >= target {
right = mid
} else {
left = mid + 1
}
}
return left
}
}
|