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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)-1
            while l<=r:
                mid=(l+r)//2
                if nums[mid]>=target:
                    r=mid-1
                else:
                    l=mid+1
            return l
            
        l=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 = nums
        let leftIndex = search(target)
        let rightIndex = search(target + 1)
        return rightIndex - leftIndex
    }

    private func search(_ x: Int) -> Int {
        var left = 0
        var right = nums.count
        while left < right {
            let mid = (left + right) / 2
            if nums[mid] >= x {
                right = mid
            } else {
                left = mid + 1
            }
        }
        return left
    }
}