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
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
}