128. 最长连续序列

Author: w2xi

README.md

@@ -84,6 +84,7 @@
 |122|[买卖股票的最佳时机 II](https://leetcode.cn/problems/best-time-to-buy-and-sell-stock-ii/)|[JavaScript](./algorithms/best-time-to-buy-and-sell-stock-ii.js)|Medium|
 |124|[二叉树中的最大路径和](https://leetcode.cn/problems/binary-tree-maximum-path-sum/)|[JavaScript](./algorithms/binary-tree-maximum-path-sum.js)|Hard|
 |125|[验证回文串](https://leetcode.cn/problems/valid-palindrome/)|[JavaScript](./algorithms/valid-palindrome.js)|Easy|
+|128|[最长连续序列](https://leetcode.cn/problems/longest-consecutive-sequence/)|[JavaScript](./algorithms/longest-consecutive-sequence.js)|Medium|
 |129|[求根节点到叶节点数字之和](https://leetcode.cn/problems/sum-root-to-leaf-numbers/)|[JavaScript](./algorithms/sum-root-to-leaf-numbers.js)|Medium|
 |131|[分割回文串](https://leetcode.cn/problems/palindrome-partitioning/)|[JavaScript](./algorithms/palindrome-partitioning.js)|Medium|
 |134|[加油站](https://leetcode.cn/problems/gas-station/)|[JavaScript](./algorithms/gas-station.js)|Medium|

algorithms/longest-consecutive-sequence.js

@@ -0,0 +1,52 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var longestConsecutive = function (nums) {
+  // 时间复杂度: O(nlogn)
+  // 空间复杂度: O(1)
+
+  if (nums.length <= 1) return nums.length;
+
+  nums.sort((a, b) => a - b);
+  let maxLength = 1;
+  let count = 1;
+
+  // 0 0 1 2 3 4 5 6 7 8
+
+  for (let i = 1; i < nums.length; i++) {
+    if (nums[i] === nums[i - 1] + 1) {
+      count++;
+    } else if (nums[i] !== nums[i - 1]) {
+      count = 1;
+    }
+    maxLength = Math.max(count, maxLength);
+  }
+
+  return maxLength;
+};
+
+// 时间复杂度: O(longn)
+// 空间复杂度: O(n)
+var longestConsecutive = function (nums) {
+  // 题目要求时间复杂度为 O(n), 就用空间换时间
+  // (一般在leetcode中，对时间复杂度有要求，就用空间来换，对空间复杂度有要求，就用时间来换)
+
+  const set = new Set(nums);
+  let maxLength = 0;
+
+  for (let i = 0; i < nums.length; i++) {
+    if (!set.has(nums[i] - 1)) {
+      let val = nums[i];
+      let curLen = 1;
+
+      while (set.has(val + 1)) {
+        val++;
+        curLen++;
+      }
+      maxLength = Math.max(curLen, maxLength);
+    }
+  }
+
+  return maxLength;
+};