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Author: assassint2017

Longest Increasing Subsequence/Longest_Increasing_Subsequence.cpp

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+// 典型的动态规划
+// 时间复杂度为n2的版本
+
+// Runtime: 48 ms, faster than 56.04% of C++ online submissions for Longest Increasing Subsequence.
+// Memory Usage: 8.7 MB, less than 40.00% of C++ online submissions for Longest Increasing Subsequence.
+
+class Solution 
+{
+public:
+    int lengthOfLIS(vector<int>& nums) 
+    {
+        if (nums.size() <= 1)
+            return nums.size();
+        
+        int *res = new int[nums.size()];
+        for (int i = 0; i < nums.size(); i++)
+            res[i] = 1;
+        
+        // res[i]代表以a[i]结尾的LIS的长度
+        for (int i = 1; i < nums.size(); i++)
+        {
+            for (int j = i - 1; j >= 0; j--)
+            {
+                if (nums[i] > nums[j])
+                    res[i] = max(res[i], res[j] + 1);
+            }
+        }
+        
+        // 遍历求取一下最大值
+        int maxLength = INT_MIN;
+        for (int i = 0; i < nums.size(); i++)
+            maxLength = max(maxLength, res[i]);
+        delete[] res;
+        return maxLength;
+    }
+};
+
+// 时间复杂度为nlogn的版本
+// https://leetcode.com/problems/longest-increasing-subsequence/discuss/74824/JavaPython-Binary-search-O(nlogn)-time-with-explanation
+
+// Runtime: 4 ms, faster than 100.00% of C++ online submissions for Longest Increasing Subsequence.
+// Memory Usage: 8.7 MB, less than 48.95% of C++ online submissions for Longest Increasing Subsequence.
+
+class Solution 
+{
+public:
+    int lengthOfLIS(vector<int>& nums) 
+    {
+        if (nums.size() <= 1)
+            return nums.size();
+
+        vector<int> tails;
+        tails.push_back(nums[0]);
+        for (int i = 1; i < nums.size(); i++)     
+        {
+            if (nums[i] > tails.back())
+                tails.push_back(nums[i]);
+            else
+            {
+                // 使用二分法确定需要修改的索引
+                int leftPtr = 0, rightPtr = tails.size() - 1;
+                while (leftPtr < rightPtr)
+                {
+                    int midPtr = (leftPtr + rightPtr) / 2;
+                    if (tails[midPtr] < nums[i])
+                        leftPtr = midPtr + 1;
+                    else
+                        rightPtr = midPtr;
+                }
+                tails[leftPtr] = nums[i];
+            }
+        }
+        return tails.size();
+    }
+};

Longest Increasing Subsequence/Longest_Increasing_Subsequence.py

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+# nlogn 的py代码
+
+# Runtime: 32 ms, faster than 100.00% of Python3 online submissions for Longest Increasing Subsequence.
+# Memory Usage: 12.5 MB, less than 93.78% of Python3 online submissions for Longest Increasing Subsequence.
+
+class Solution:
+    def lengthOfLIS(self, nums: 'List[int]') -> 'int':
+        if len(nums) <= 1:
+            return len(nums)
+        
+        tails = []
+        tails.append(nums[0])
+
+        for i in range(1, len(nums)):
+            if nums[i] > tails[-1]:
+                tails.append(nums[i])
+            else:
+                leftPtr, rightPtr = 0, len(tails) - 1
+                while leftPtr < rightPtr:
+                    midPtr = (leftPtr + rightPtr) // 2
+                    if tails[midPtr] < nums[i]:
+                        leftPtr = midPtr + 1
+                    else:
+                        rightPtr = midPtr
+                tails[leftPtr] = nums[i]
+        
+        return len(tails)