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Author: assassint2017

Length of Longest Fibonacci Subsequence/Length_of_Longest_Fibonacci_Subsequence.cpp

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+// 第一种比较慢的思路
+// Runtime: 292 ms, faster than 19.15% of C++ online submissions for Length of Longest Fibonacci Subsequence.
+// Memory Usage: 9.5 MB, less than 85.71% of C++ online submissions for Length of Longest Fibonacci Subsequence.
+
+class Solution 
+{
+public:
+    int lenLongestFibSubseq(vector<int>& input) 
+    {
+        // 边界条件处理
+        if (input.size() <= 2)
+            return input.size();
+
+        unordered_set<int> hashset(input.begin(), input.end());
+
+        int maxLength = 0;
+        for (int i = 0; i < input.size(); ++i)
+        {
+            for (int j = i + 1; j < input.size(); ++j)
+            {
+                int curLength = 2;
+
+                int a = input[i];
+                int b = input[j];
+
+                while (hashset.find(a + b) != hashset.end())
+                {
+                    b = a + b;
+                    a = b - a;
+                    ++curLength;
+                }
+        
+                maxLength = max(maxLength, curLength);
+            }
+        }
+        return maxLength >= 3 ? maxLength : 0;
+    }
+};
+
+
+
+// 第二种思路，使用动态规划进行加速 dp[i][j] 代表着以索引为i和j的元素为最后两位的斐波那契数列的最大长度
+// Runtime: 252 ms, faster than 23.11% of C++ online submissions for Length of Longest Fibonacci Subsequence.
+// Memory Usage: 62.4 MB, less than 14.29% of C++ online submissions for Length of Longest Fibonacci Subsequence.
+class Solution
+{
+public:
+	int lenLongestFibSubseq(vector<int>& input)
+	{
+		// 边界条件处理
+		if (input.size() <= 2) return 0;
+
+		int size = input.size();
+		vector<vector<int>> dp(size, vector<int>(size, 2));
+		unordered_map<int, int> hashmap;
+
+		int maxLength = 0;
+		for (int j = 0; j < input.size(); ++j)
+		{
+			hashmap.insert(pair<int, int>(input[j], j));
+			for (int i = 0; i < j; ++i)
+			{
+				int a = input[i], b = input[j];
+
+				if (hashmap.find(b - a) != hashmap.end() && hashmap.at(b - a) < i)
+					dp[i][j] = max(dp[hashmap.at(b - a)][i] + 1, dp[i][j]);
+
+				maxLength = max(maxLength, dp[i][j]);
+			}
+		}
+		return maxLength >= 3 ? maxLength : 0;
+	}
+};