題目

https://leetcode.com/problems/longest-consecutive-sequence/

題解

方法1：HashMap 解法，O(n^2)

如下圖，假設 n=4 被插入，count 的變化如下：

class Solution {public int longestConsecutive(int[] nums) {HashMap<Integer, Integer> dp = new HashMap<>();for (int n : nums) {if (!dp.containsKey(n)) {int count = 1; // 以n結尾的最長連續序列長度if (dp.containsKey(n - 1)) {count += dp.get(n - 1); // dp(n)=dp(n-1)+1}dp.put(n, count);// 比n大的連續key，將value依次累加dp(n)int next = n + 1;while (dp.containsKey(next)) {dp.put(next, dp.get(next) + count);next++;}} // else, duplicate, do nothing}int max = 0;for (int k : dp.keySet()) {max = Math.max(max, dp.get(k));}return max;} }

方法2：官方題解，O(n)

https://leetcode.com/problems/longest-consecutive-sequence/solution/

class Solution {public int longestConsecutive(int[] nums) {Set<Integer> num_set = new HashSet<Integer>();for (int num : nums) {num_set.add(num);}int longestStreak = 0;for (int num : num_set) {if (!num_set.contains(num-1)) {int currentNum = num;int currentStreak = 1;while (num_set.contains(currentNum+1)) {currentNum += 1;currentStreak += 1;}longestStreak = Math.max(longestStreak, currentStreak);}}return longestStreak;} }

方法3：dp

如果 num 的加入使連續序列的長度增加，那么一定是因為它與 num-1 或 num+1 連成了一片。

連成一片后，只需要讓連續序列的兩個端點保持最新記錄，不需要關心除端點以外的元素。

class Solution:def longestConsecutive(self, nums: List[int]) -> int:result = 0record = {}for num in nums:if num in record: continuel, r = num, numif num - 1 in record:l = record[num - 1][0]if num + 1 in record:r = record[num + 1][1]if r - l + 1 > result:result = r - l + 1record[num] = (l, r) # 占位，避免重復record[l] = (l, r)record[r] = (l, r)for r in record:print(str(r) + ": " + str(record[r]))return result

總結

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