【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

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
}

原文地址：https://blog.csdn.net/shuting7/article/details/136450108

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