[add] LeetCode 128. Longest Consecutive Sequence

Author: brianway

algorithms-leetcode/README.md

@@ -71,6 +71,7 @@ Leetcode problems classified by company:
 |120|Triangle|Medium|||
 |121|Best Time to Buy and Sell Stock|Easy||待查最优解，以及一题多解|
 |122|Best Time to Buy and Sell Stock II|Medium|Dynamic Programming/Greedy|一题多解|
+|128|Longest Consecutive Sequence|Medium|Hash Table|TODO一题多解|
 |139|Word Break|Medium|Dynamic Programming,完全背包||
 |141|Linked List Cycle|Easy|Two Pointers||
 |144|Binary Tree Preorder Traversal|Easy|Binary Tree/Stack|一题多解|

algorithms-leetcode/src/main/java/com/brianway/learning/algorithms/leetcode/medium/LongestConsecutiveSequence.java

@@ -0,0 +1,57 @@
+package com.brianway.learning.algorithms.leetcode.medium;
+
+import java.util.HashSet;
+
+/**
+ * LeetCode 128. Longest Consecutive Sequence
+ * Question: https://leetcode.com/problems/longest-consecutive-sequence/
+ * 关键题设：runs in O(n) time.
+ *
+ * @auther brian
+ * @since 2022/9/6 23:59
+ */
+public class LongestConsecutiveSequence {
+    public int longestConsecutive(int[] nums) {
+        return 0;
+    }
+
+    /**
+     * 哈希表
+     * <p>
+     * 先构建哈希集合，用于去重；
+     * 再依次遍历其中的元素num：只统计num-1不在其中时，一直next,看nums++是否在哈希集合中，统计next的次数
+     * 1）如果num-1在里面，则忽略跳过
+     * 2)如果num-1不在里面，则每次查看看num是否在里面
+     *
+     * <p>
+     * 时间复杂度：O(n)
+     * 空间复杂度：O(n)
+     */
+    public class LongestConsecutiveSequence0 extends LongestConsecutiveSequence {
+        @Override
+        public int longestConsecutive(int[] nums) {
+            // 构建哈希Set
+            HashSet<Integer> hashSet = new HashSet<>();
+            for (int i = 0; i < nums.length; i++) {
+                hashSet.add(nums[i]);
+            }
+            int maxCount = 0;
+            // 遍历
+            for (int num : hashSet) {
+                if (!hashSet.contains(num - 1)) {
+                    // 进入统计循环
+                    int curCount = 1;
+                    int cur = num;
+                    while (hashSet.contains(++cur)) {
+                        curCount++;
+                    }
+                    maxCount = Math.max(maxCount, curCount);
+                }
+                // else 忽略
+            }
+            return maxCount;
+        }
+    }
+
+    // TODO 其他解法
+}