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2006. 差的绝对值为 K 的数对数目

English Version

题目描述

给你一个整数数组 nums 和一个整数 k ，请你返回数对 (i, j) 的数目，满足 i < j 且 |nums[i] - nums[j]| == k 。

|x| 的值定义为：

• 如果 x >= 0 ，那么值为 x 。
• 如果 x < 0 ，那么值为 -x 。

 

示例 1：

输入：nums = [1,2,2,1], k = 1
输出：4
解释：差的绝对值为 1 的数对为：
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]

示例 2：

输入：nums = [1,3], k = 3
输出：0
解释：没有任何数对差的绝对值为 3 。

示例 3：

输入：nums = [3,2,1,5,4], k = 2
输出：3
解释：差的绝对值为 2 的数对为：
- [3,2,1,5,4]
- [3,2,1,5,4]
- [3,2,1,5,4]

 

提示：

• 1 <= nums.length <= 200
• 1 <= nums[i] <= 100
• 1 <= k <= 99

解法

方法一：暴力法

由于 nums 长度范围是 [1,200]，故可以直接双重循环遍历计数。

方法二：哈希表计数，一次遍历

Python3

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        ans, n = 0, len(nums)
        for i in range(n):
            for j in range(i + 1, n):
                if abs(nums[i] - nums[j]) == k:
                    ans += 1
        return ans

class Solution:
    def countKDifference(self, nums: List[int], k: int) -> int:
        ans = 0
        counter = Counter()
        for num in nums:
            ans += counter[num - k] + counter[num + k]
            counter[num] += 1
        return ans

Java

class Solution {
    public int countKDifference(int[] nums, int k) {
        int ans = 0;
        for (int i = 0, n = nums.length; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (Math.abs(nums[i] - nums[j]) == k) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

class Solution {
    public int countKDifference(int[] nums, int k) {
        int ans = 0;
        int[] counter = new int[110];
        for (int num : nums) {
            if (num >= k) {
                ans += counter[num - k];
            }
            if (num + k <= 100) {
                ans += counter[num + k];
            }
            ++counter[num];
        }
        return ans;
    }
}

TypeScript

function countKDifference(nums: number[], k: number): number {
    let ans = 0;
    let cnt = new Map();
    for (let num of nums) {
        ans += (cnt.get(num - k) || 0) + (cnt.get(num + k) || 0);
        cnt.set(num, (cnt.get(num) || 0) + 1);
    }
    return ans;
}

C++

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int n = nums.size();
        int ans = 0;
        for (int i = 0; i < n; ++i)
            for (int j = i + 1; j < n; ++j)
                if (abs(nums[i] - nums[j]) == k) ++ans;
        return ans;
    }
};

class Solution {
public:
    int countKDifference(vector<int>& nums, int k) {
        int ans = 0;
        vector<int> counter(110);
        for (int num : nums)
        {
            if (num >= k) ans += counter[num - k];
            if (num + k <= 100) ans += counter[num + k];
            ++counter[num];
        }
        return ans;
    }
};

Go

func countKDifference(nums []int, k int) int {
	n := len(nums)
	ans := 0
	for i := 0; i < n; i++ {
		for j := i + 1; j < n; j++ {
			if abs(nums[i]-nums[j]) == k {
				ans++
			}
		}
	}
	return ans
}

func abs(x int) int {
	if x > 0 {
		return x
	}
	return -x
}

func countKDifference(nums []int, k int) int {
	ans := 0
	counter := make([]int, 110)
	for _, num := range nums {
		if num >= k {
			ans += counter[num-k]
		}
		if num+k <= 100 {
			ans += counter[num+k]
		}
		counter[num]++
	}
	return ans
}

Rust

impl Solution {
    pub fn count_k_difference(nums: Vec<i32>, k: i32) -> i32 {
        let mut res = 0;
        let n = nums.len();
        for i in 0..n - 1 {
            for j in i..n {
                if (nums[i] - nums[j]).abs() == k {
                    res += 1;
                }
            }
        }
        res
    }
}

impl Solution {
    pub fn count_k_difference(nums: Vec<i32>, k: i32) -> i32 {
        let mut arr = [0; 101];
        let mut res = 0;
        for num in nums {
            if num - k >= 1 {
                res += arr[(num - k) as usize];
            }
            if num + k <= 100 {
                res += arr[(num + k) as usize]
            }
            arr[num as usize] += 1;
        }
        res
    }
}

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