674_Longest_Continuous_Increasing_Subsequence

qiyuangong · qiyuangong · commit 5a75da2ea3f3 · 2018-12-23T13:50:52.000+08:00
diff --git a/README.md b/README.md
@@ -163,6 +163,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 617 | [Merge Two Binary Trees](https://leetcode.com/problems/merge-two-binary-trees/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/617_Merge_Two_Binary_Trees.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/617_Merge_Two_Binary_Trees.java) | Traverse both trees Recursion & Iterative (stack) |
 | 628 | [Maximum Product of Three Numbers](https://leetcode.com/problems/maximum-product-of-three-numbers/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/628_Maximum_Product_of_Three_Numbers.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/628_Maximum_Product_of_Three_Numbers.java) | Actually, we should only care about min1, min2 and max1-max3, to find these five elements, we can use 1. Brute force, O(n^3) and O(1)<br>2. Sort, O(nlogn) and O(1)<br>3. Single scan with 5 elements keep, O(n) and O(1) |
 | 654 | [Maximum Binary Tree](https://leetcode.com/problems/maximum-binary-tree/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/654_Maximum_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/654_Maximum_Binary_Tree.java) | 1. Divide and conquer, recursive, O(n^2)<br>2. Monotonic stack, O(n) |
+| 674 | [Longest Continuous Increasing Subsequence](https://leetcode.com/problems/longest-continuous-increasing-subsequence/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/674_Longest_Continuous_Increasing_Subsequence.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/674_Longest_Continuous_Increasing_Subsequence.java) | Scan nums once, check nums[i] < nums[i+1], if not reset count, O(n) and O(1) |
 | 680 | [Valid Palindrome II](https://leetcode.com/problems/valid-palindrome-ii/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/680_Valid_Palindrome_II.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/680_Valid_Palindrome_II.java) | Recursively check s[left == end, when not equal delete left or right. |
 | 692 | [Top K Frequent Words](https://leetcode.com/problems/top-k-frequent-words/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/692_Top_K_Frequent_Words.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/692_Top_K_Frequent_Words.java) | 1. Sort based on frequency and alphabetical order, O(nlgn) and O(n)<br>2. Find top k with Heap, O(nlogk) and O(n) |
 | 695 | [Max Area of Island](https://leetcode.com/problems/max-area-of-island/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/695_Max_Area_of_Island.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/695_Max_Area_of_Island.java) | 1. DFS, O(n^2) and O(n)<br>2. BFS, O(n^2) and O(n)|
diff --git a/java/674_Longest_Continuous_Increasing_Subsequence.java b/java/674_Longest_Continuous_Increasing_Subsequence.java
@@ -0,0 +1,15 @@
+class Solution {
+    public int findLengthOfLCIS(int[] nums) {
+        if (nums.length == 0) return 0;
+        int curr = 1, ans = 1;
+        for (int i = 0; i < nums.length - 1; i++) {
+            if (nums[i] < nums[i + 1]) {
+                curr ++;
+                if (curr >= ans) ans = curr;
+            } else {
+                curr = 1;
+            }
+        }
+        return ans;
+    }
+}
diff --git a/python/674_Longest_Continuous_Increasing_Subsequence.py b/python/674_Longest_Continuous_Increasing_Subsequence.py
@@ -0,0 +1,17 @@
+class Solution(object):
+    def findLengthOfLCIS(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        if not nums or len(nums) == 0:
+            return 0
+        ans = curr = 1
+        for i in range(len(nums) - 1):
+            if nums[i] < nums[i + 1]:
+                curr += 1
+                if curr >= ans:
+                    ans = curr
+            else:
+                curr = 1
+        return ans
