Merge branch 'master' of https://github.com/azl397985856/leetcode

lucifer · lucifer · commit 9f0ad1cdad81 · 2020-12-30T16:01:57.000+08:00
diff --git a/problems/128.longest-consecutive-sequence.md b/problems/128.longest-consecutive-sequence.md
@@ -71,6 +71,53 @@ return Math.max(count, maxCount);
 
 ## 代码
 
+代码支持：Java, Python，JS
+
+Java Code:
+
+```java
+class Solution {
+    public int longestConsecutive(int[] nums) {
+        Set<Integer> set = new HashSet<Integer>();
+        int ans = 0;
+        for (int num : nums) {
+            set.add(num);
+        }
+        for(int i = 0;i < nums.length; i ++) {
+            int x = nums[i];
+            // 说明x是连续序列的开头元素
+            if  (!set.contains(x - 1)) {
+                while(set.contains(x + 1)) {
+                    x ++;
+                }
+            }
+            ans = Math.max(ans, x - nums[i] + 1);
+        }
+        return ans;
+        
+    }
+}
+```
+
+Python Code:
+
+```py
+class Solution:
+    def longestConsecutive(self, A: List[int]) -> int:
+        seen = set(A)
+        ans = 0
+        for a in A:
+            t = a
+            #  if 的作用是剪枝
+            if t + 1 not in seen:
+                while t - 1 in seen:
+                    t -= 1
+            ans = max(ans, a - t + 1)
+        return ans
+```
+ 
+ JS Code:
+
 ```js
 /**
  * @param {number[]} nums
@@ -81,7 +128,7 @@ var longestConsecutive = function(nums) {
   let max = 0;
   let temp = 0;
   set.forEach(x => {
-    // 说明x是连续序列的开头元素
+    // 说明x是连续序列的开头元素。加这个条件相当于剪枝的作用，否则时间复杂度会退化到 N ^ 2
     if (!set.has(x - 1)) {
       temp = x + 1;
       while (set.has(y)) {
