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面试题 53 - I. 在排序数组中查找数字 I

题目描述

统计一个数字在排序数组中出现的次数。

示例 1:

输入: nums = [5,7,7,8,8,10], target = 8
输出: 2

示例 2:

输入: nums = [5,7,7,8,8,10], target = 6
输出: 0

提示：

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums 是一个非递减数组
• -109 <= target <= 109

注意：本题与主站 34 题相同（仅返回值不同）：https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/

解法

方法一：二分查找

由于数组 nums 已排好序，我们可以使用二分查找的方法找到数组中第一个大于等于 target 的元素的下标 $l$，以及第一个大于 target 的元素的下标 $r$，那么 target 的个数就是 $r-l$。

时间复杂度 $O(\log n)$，其中 $n$ 为数组的长度。空间复杂度 $O(1)$。

【二分查找 红蓝染色法】 https://www.bilibili.com/video/BV1AP41137w7/?share_source=copy_web&vd_source=ed4a51d52f6e5c9a2cb7def6fa64ad6a

Python3

class Solution:def search(self, nums: List[int], target: int) -> int:#l = bisect_left(nums, target)#r = bisect_right(nums, target)def bi_sect(nums,target):l,r=0,len(nums)-1while l<=r:mid=(l+r)//2if nums[mid]>=target:r=mid-1else:l=mid+1return ll=bi_sect(nums,k)r=bi_sect(nums,k+1)return r-l

Java

class Solution {private int[] nums;public int search(int[] nums, int target) {this.nums = nums;int l = search(target);int r = search(target + 1);return r - l;}private int search(int x) {int l = 0, r = nums.length;while (l < r) {int mid = (l + r) >>> 1;if (nums[mid] >= x) {r = mid;} else {l = mid + 1;}}return l;}
}

C++

class Solution {
public:int search(vector<int>& nums, int target) {auto l = lower_bound(nums.begin(), nums.end(), target);auto r = upper_bound(nums.begin(), nums.end(), target);return r - l;}
};

Go

func search(nums []int, target int) int {l := sort.SearchInts(nums, target)r := sort.SearchInts(nums, target+1)return r - l
}

Rust

impl Solution {pub fn search(nums: Vec<i32>, target: i32) -> i32 {let search = |x| {let mut l = 0;let mut r = nums.len();while l < r {let mid = l + (r - l) / 2;if nums[mid] >= x {r = mid;} else {l = mid + 1;}}l as i32};search(target + 1) - search(target)}
}

JavaScript

/*** @param {number[]} nums* @param {number} target* @return {number}*/
var search = function (nums, target) {const search = x => {let l = 0;let r = nums.length;while (l < r) {const mid = (l + r) >> 1;if (nums[mid] >= x) {r = mid;} else {l = mid + 1;}}return l;};const l = search(target);const r = search(target + 1);return r - l;
};

C#

public class Solution {public int Search(int[] nums, int target) {int l = search(nums, target);int r = search(nums, target + 1);return r - l;}private int search(int[] nums, int x) {int l = 0, r = nums.Length;while (l < r) {int mid = (l + r) >> 1;if (nums[mid] >= x) {r = mid;} else {l = mid + 1;}}return l;}
}

Swift

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