LeetCode //C - 219. Contains Duplicate II

219. Contains Duplicate II

Given an integer array nums and an integer k , return true if there are two distinct indices i and j in the array such that numsi == numsj and abs(i - j) <= k .

Example 1:

Input: nums = 1,2,3,1, k = 3
Output: true

Example 2:

Input: nums = 1,0,1,1, k = 1
Output: true

Example 3:

Input: nums = 1,2,3,1,2,3, k = 2
Output: false

Constraints:

  • 1 < = n u m s . l e n g t h < = 1 0 5 1 <= nums.length <= 10^5 1<=nums.length<=105
  • − 1 0 9 < = n u m s i < = 1 0 9 -10^9 <= numsi <= 10^9 −109<=numsi<=109
  • 0 < = k < = 1 0 5 0 <= k <= 10^5 0<=k<=105

From: LeetCode

Link: 219. Contains Duplicate II


Solution:

Ideas:
  1. Structure Definition: A structure Number is defined to store two integers: the value of an element from the nums array and its original index in the array.

  2. Array Initialization: An array numbers of type Number is created and initialized with the values and corresponding indices from the input array nums.

  3. Sorting: The numbers array is sorted by the values using the qsort function. The compare function is used to define the sorting order, which is based on the value field of the structure. After sorting, elements with the same value will be adjacent to each other, and their original indices can still be accessed.

  4. Scanning for Duplicates: The sorted numbers array is then scanned to find adjacent pairs with the same value. If such a pair is found, the original indices are checked to see if the absolute difference between them is less than or equal to k. If this condition is met, the function returns true.

  5. Memory Cleanup: The dynamically allocated memory for the numbers array is freed before exiting the function.

  6. Result: If no such pair is found that meets the condition, the function returns false.

By sorting the numbers and keeping track of the original indices, this approach efficiently checks for duplicates within the specified range k. The sorting ensures that the same values are grouped together, making it easy to check the condition without needing a hash table.

Code:
c 复制代码
typedef struct {
    int value;
    int index;
} Number;

int compare(const void *a, const void *b) {
    return ((Number*)a)->value - ((Number*)b)->value;
}

bool containsNearbyDuplicate(int* nums, int numsSize, int k) {
    if (k == 0 || numsSize <= 1) return false;
    
    Number* numbers = (Number*)malloc(numsSize * sizeof(Number));
    for (int i = 0; i < numsSize; i++) {
        numbers[i].value = nums[i];
        numbers[i].index = i;
    }
    
    qsort(numbers, numsSize, sizeof(Number), compare);
    
    for (int i = 0; i < numsSize - 1; i++) {
        if (numbers[i].value == numbers[i + 1].value && 
            abs(numbers[i].index - numbers[i + 1].index) <= k) {
            free(numbers);
            return true;
        }
    }
    
    free(numbers);
    return false;
}
相关推荐
炸膛坦客11 分钟前
单片机/C/C++八股:(二十六)IIC 专题(I²C)---- 上集
c语言·c++·单片机
旖-旎26 分钟前
LeetCode 518:零钱兑换||(完全背包)—— 题解
c++·算法·leetcode·动态规划·背包问题
To_OC1 小时前
LC 42 接雨水:暴力超时卡半天?前后缀数组一用就通了
javascript·算法·leetcode
delishcomcn2 小时前
AI视觉识别+分切算法:电化铝缺陷检测与裁切一体化解锁
人工智能·算法
触底反弹2 小时前
深入理解大模型采样:Temperature、Top-K、Top-P 的原理与实战
人工智能·算法·面试
雪碧聊技术2 小时前
力扣 LCR 091. 粉刷房子 —— 动态规划入门详解
算法·动态规划
CV-Climber5 小时前
检索技术的实际应用
人工智能·算法
hhzz6 小时前
Tiger AI Platform平台中增加人脸识别功能
图像处理·人工智能·算法·计算机视觉·大模型
从零开始的代码生活_6 小时前
C++ 继承详解:访问控制、对象模型、菱形继承与设计取舍
开发语言·c++·后端·学习·算法