剑指 Offer II 068. 查找插入位置

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剑指 Offer II 068. 查找插入位置

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

Python3

import bisect as bi
class Solution:def searchInsert(self, nums: List[int], target: int) -> int:pos=bi.bisect_left(nums,target)return posclass Solution:def searchInsert(self, nums: List[int], target: int) -> int:left, right = 0, len(nums)-1while left <=right:mid = (left + right) >> 1if nums[mid] >= target:right = mid-1else:left = mid + 1return left

Java

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;}
}

C++

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;elseleft = mid + 1;}return left;}
};

Go

func searchInsert(nums []int, target int) int {left, right := 0, len(nums)for left < right {mid := (left + right) >> 1if nums[mid] >= target {right = mid} else {left = mid + 1}}return left
}

JavaScript

/*** @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;
};

Swift

class Solution {func searchInsert(_ nums: [Int], _ target: Int) -> Int {var left = 0var right = nums.countwhile left < right {let mid = (left + right) / 2if nums[mid] >= target {right = mid} else {left = mid + 1}}return left}
}