cxxpython9
diff --git a/‎.idea/workspace.xml‎
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diff --git a/‎leetcode/300. Longest Increasing Subsequence.py‎
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+# -*- coding:utf-8 -*-
+'''
+Given an unsorted array of integers, find the length of longest increasing subsequence.
+
+For example,
+Given [10, 9, 2, 5, 3, 7, 101, 18],
+The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
+
+Your algorithm should run in O(n^2) complexity.
+
+Follow up: Could you improve it to O(n log n) time complexity?
+'''
+
+
+class Solution(object):
+    # 原始O(nlogn)方法
+    def lengthOfLIS(self, nums):
+        tails = [0] * len(nums)
+        size = 0
+        for x in nums:
+            i, j = 0, size
+            while i != j:
+                mid = (i + j) // 2
+                if tails[mid] < x:
+                    i = mid + 1
+                else:
+                    j = mid
+            tails[i] = x
+            size = max(i + 1, size)
+        return size
+    # 简单优化的O(nlogn)方法, 1000000长度为8的数据平均节约0.5s
+    def lengthOfLIS2(self, nums):
+        tails = [0] * len(nums)
+        size = 0
+        for x in nums:
+            i, j = 0, size
+            if j > 0 and tails[j - 1] < x:
+                tails[j] = x
+                size += 1
+                continue
+            while i != j:
+                mid = (i + j) // 2
+                if tails[mid] < x:
+                    i = mid + 1
+                else:
+                    j = mid
+            tails[i] = x
+            size = max(i + 1, size)
+        return size
+
+s = Solution()
+aList = [3, 2, 4, 9, 10, 22, 15, 17, 23, 6]
+print(s.lengthOfLIS(aList), s.lengthOfLIS2(aList))