【力扣 - 最长连续数组】

题目描述

给定一个未排序的整数数组 nums ,找出数字连续的最长序列(不要求序列元素在原数组中连续)的长度。

请你设计并实现时间复杂度为 O(n) 的算法解决此问题。

示例 1:

输入:nums = [100,4,200,1,3,2]

输出:4

解释:最长数字连续序列是 [1, 2, 3, 4]。它的长度为 4。

示例 2:

输入:nums = [0,3,7,2,5,8,4,6,0,1]

输出:9

提示:

0 <= nums.length <= 10^5

-10^9 <= nums[i] <= 10^9

题解

思路

先排序再比较计数;

如果相邻两数差一计数,如果相等进入下一层循环判断;

如果后面数与前一个数既不相等又不比前一个多一,重值计数为1.

代码

c 复制代码
int cmp(const void* a, const void* b)
{
    // Custom comparison function for qsort to sort integers in ascending order
    return (long long)*(int*)a - (long long)*(int*)b;
}

int longestConsecutive(int* a, int n)
{
    // Check for edge cases
    if (a == NULL || n == 0) {
        return 0;
    }
    
    /* The  qsort  function in C is used to sort an array in ascending order. 
     * It takes the following parameters: 
     * base: Pointer to the array to be sorted. 
     * nmemb: Number of elements in the array. 
     * size: Size in bytes of each element in the array. 
     * compar: Pointer to a comparison function that determines the order of elements. 
     * The comparison function ( compar ) should return an integer less than, 
     * equal to, or greater than zero if the first argument is considered to be 
     * respectively less than, equal to, or greater than the second argument. 
     * void qsort(void *base, size_t nmemb, size_t size, int (*compar)(const void *, const void *));
     */
    // Sort the input array 'a' in ascending order
    qsort(a, n, sizeof(int), cmp);
    
    int t = a[0]; // Initialize 't' with the first element of the sorted array
    int cnt = 1; // Initialize 'cnt' to keep track of the current consecutive sequence length
    int max = 1; // Initialize 'max' to keep track of the maximum consecutive sequence length
    
    // Iterate through the sorted array to find the longest consecutive sequence
    for (int i = 1; i < n; i++) {
        if (a[i] == t) {
            // If the current element is equal to the previous element, skip it
            continue;
        } else if ((long long)a[i] - t == 1LL) {
            // If the current element is consecutive to the previous element
            t = a[i]; // Update 't' to the current element
            cnt++; // Increment the consecutive count
            if (max < cnt) {
                max = cnt; // Update 'max' if a longer consecutive sequence is found
            }
        } else {
            // If the current element is not consecutive to the previous element
            t = a[i]; // Update 't' to the current element
            cnt = 1; // Reset the consecutive count
        }
    }
    
    return max; // Return the maximum consecutive sequence length
}
相关推荐
潇湘夜雨6978 分钟前
2023ICPC合肥题解
算法
Small踢倒coffee_氕氘氚12 分钟前
iPhone闹钟无法识别调休致用户迟到,苹果客服称会记录反馈
笔记·算法·灌灌灌灌
制冷男孩16 分钟前
机器学习算法-支持向量机SVM
人工智能·算法·机器学习·支持向量机
代码程序猿RIP31 分钟前
【C语言干货】回调函数
c语言·开发语言·数据结构·c++·算法
刚入门的大一新生1 小时前
C++初阶-模板初阶
开发语言·数据结构·c++
mljy.1 小时前
递归、搜索和回溯算法《递归》
算法
知识漫步1 小时前
代码随想录算法训练营第60期第二十一天打卡
数据结构·算法
刃神太酷啦1 小时前
排序--数据结构初阶(4)(C/C++)
c语言·数据结构·c++·算法·leetcode·深度优先·广度优先
凯子坚持 c1 小时前
深度解析算法之分治(归并)
算法·leetcode·职场和发展
是店小二呀1 小时前
【优选算法-二分查找】二分查找算法解析:如何通过二段性优化搜索效率
c++·算法