LeetCode-Longest Consecutive Sequence 哎,这题又不会做,想了好久,还是想不出来,最长连续的数的长度,首先想到的肯定是排序,O(nlogn),不过题目要求是O(n), 于是又想到用hash的思想,类似数组记数的方式,不过即便不考虑存的空间的话,因为给定的数的大小并不在一定的范围之内, 所以最后的时间复杂度会
Longest Consecutive Sequence Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example, Given[100, 4, 200, 1, 3, 2], The longest consecutive elements sequence is[1, 2, 3, 4]. Return its length:4. Your algorithm should run in O(n) ...
实现 public int longestConsecutive(int[] nums) { if(nums.length == 0) { return 0; } // 排序 Arrays.sort(nums); int maxLen = 1; int tempLen = 1; // 对于连续的定义是什么? for(int i = 1; i < nums.length; i++) { int num = nums[i]; int pre = nums[i-1]; if(num ...
[LeetCode]Longest Consecutive Sequence 题目大意 给定一个整形数组,求出最长的连续序列。例如数组[100,4,200,1,3,2],最长的连续序列长度为[1,2,3,4],长度为4。要求时间复杂度为O(n)。 思路 "排序转换成经典的动态规划问题"的话排序至少需要时间复杂度为O(nlog(n))——pass 利用c++中的set,直接会排序,...
Can you solve this real interview question? Longest Consecutive Sequence - Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time. Example 1: Inp
Given an unsorted array of integersnums, returnthe length of the longest consecutive elements sequence. You must write an algorithm that runs inO(n)time. Example 1: [1, 2, 3, 4] Example 2: Input: nums = [0,3,7,2,5,8,4,6,0,1] ...
力扣.128 最长连续序列 longest-consecutive-sequence 力扣.1 两数之和 N 种解法 two-sum 力扣.167 两数之和 II two-sum-ii 力扣.170 两数之和 III two-sum-iii 力扣.653 两数之和 IV two-sum-IV 力扣.015 三数之和 three-sum 题目 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...
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.You must write an algorithm that runs in O(n) time. Example 1:Input: nums = [100,4,200,1,3,2]Output: 4Explanation: The longest consecutive elements s
public int longestConsecutive(int[] nums) { if(nums.length == 0) { return 0; } // 排序 Arrays.sort(nums); int maxLen = 1; int tempLen = 1; // 对于连续的定义是什么? for(int i = 1; i < nums.length; i++) { int num = nums[i]; int pre = nums[i-1]; if(num - pre...
int longestConsecutive(vector<int>& nums) { if (nums.empty()) return 0; set<int> s(nums.begin(), nums.end()); int res = 1; int count = 1; set<int>::iterator first = s.begin(); set<int>::iterator last = first++; while (last != s.end()) { if (*first+1 == *last...