Add 509_Fibonacci_Number

Author: qiyuangong

README.md

@@ -154,6 +154,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 475 | [Heaters](https://leetcode.com/problems/heaters/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/475_Heaters.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/475_Heaters.java) | 1. Binary search hourse in heater array, O(nlogn) and O(1)<br> 2. Two points, O(nlogn) and O(1) |
 | 479 | [Largest Palindrome Product](https://leetcode.com/problems/largest-palindrome-product/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/479_Largest_Palindrome_Product.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/479_Largest_Palindrome_Product.java) | 1. Product max palindrome than check, O(n^2) and O(1)<br>2. [Math](# https://leetcode.com/problems/largest-palindrome-product/discuss/96305/Python-Solution-Using-Math-In-48ms) |
 | 482 | [License Key Formatting](https://leetcode.com/problems/license-key-formatting/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/482_License_Key_Formatting.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/482_License_Key_Formatting.java) | String processing, lower and len % K, O(n) and O(n) |
+| 509 | [Fibonacci Number](https://leetcode.com/problems/fibonacci-number/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/509_Fibonacci_Number.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/509_Fibonacci_Number.java) | 1. Recursive, O(n) <br>2. DP with memo, O(n). Note that N<=30, which means that we can keep a memo from 0 to 30. |
 | 538 | [Convert BST to Greater Tree](https://leetcode.com/problems/convert-bst-to-greater-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/538_Convert_BST_to_Greater_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/538_Convert_BST_to_Greater_Tree.java) | Right first DFS with a variable recording sum of node.val and right.val. 1. Recursive.<br>2. Stack 3. Reverse Morris In-order Traversal |
 | 543 | [Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/description/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/543_Diameter_of_Binary_Tree.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/543_Diameter_of_Binary_Tree.java) | DFS with O(1) for max answer |
 | 547 | [Friend Circles](https://leetcode.com/problems/friend-circles/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/547_Friend_Circles.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/547_Friend_Circles.java) | 1. DFS, O(n^2) and O(1)<br>2. BFS, O(n^2) and O(1)<br>3. Union-find, O(n^2) and O(n)|
@@ -205,7 +206,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 953 | [Verifying an Alien Dictionary](https://leetcode.com/contest/weekly-contest-114/problems/verifying-an-alien-dictionary/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/953_Verifying_an_Alien_Dictionary.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/953_Verifying_an_Alien_Dictionary.java) | Use hashmap to store index of each value, then create a comparator based on this index, O(n) and O(n) |
 | 954 | [Array of Doubled Pairs](https://leetcode.com/contest/weekly-contest-114/problems/array-of-doubled-pairs/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/954_Array_of_Doubled_Pairs.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/954_Array_of_Doubled_Pairs.java) | Sort, then use hashmap to store the frequency of each value. Then, check n, 2 * n in hashmap, O(nlogn) and O(n) |
 | 961 | [N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/submissions/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/961_N-Repeated_Element_in_Size_2N_Array.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/961_N-Repeated_Element_in_Size_2N_Array.java) | Hash and count number, O(n) and O(n) |
-| 962 | [Maximum Width Ramp](https://leetcode.com/problems/maximum-width-ramp/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/962_Maximum_Width_Ramp.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/962_Maximum_Width_Ramp.java) | 1. Sort index by value, then transfer problem into finding max gap between index, O(nlogn) and O(1)<br> |
+| 962 | [Maximum Width Ramp](https://leetcode.com/problems/maximum-width-ramp/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/962_Maximum_Width_Ramp.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/962_Maximum_Width_Ramp.java) | 1. Sort index by value, then transfer problem into finding max gap between index, O(nlogn) and O(1)<br>2. Binary Search for candicates, O(nlogn) and O(n) |
 
 | # | To Understand |
 |---| ----- |

java/509_Fibonacci_Number.java

@@ -0,0 +1,25 @@
+class Solution {
+    /*public int fib(int N) {
+        // Recursively, O(n)
+        if (N == 0) return 0;
+        if (N == 1) return 1;
+        return fib(N - 1) + fib(N - 2);
+    }*/
+
+    private List<Integer> memo;
+
+    public Solution() {
+        memo = new ArrayList();
+        memo.add(0);
+        memo.add(1);
+    }
+
+    public int fib(int N) {
+        // Dp with memo, O(n)
+        if (N < memo.size()) return memo.get(N);
+        for (int i = memo.size(); i <= N; i++) {
+            memo.add(memo.get(i - 1) + memo.get(i - 2));
+        }
+        return memo.get(N);
+    }
+}

python/509_Fibonacci_Number.py

@@ -0,0 +1,30 @@
+class Solution(object):
+
+    def __init__(self):
+        self.memo = []
+        self.memo.append(0)
+        self.memo.append(1)
+
+    def fib(self, N):
+        """
+        DP with memo
+        :type N: int
+        :rtype: int
+        """
+        if N < len(self.memo):
+            return self.memo[N]
+        for i in range(len(self.memo), N + 1):
+            self.memo.append(self.memo[i - 1] + self.memo[i - 2])
+        return self.memo[N]
+
+    # def fib(self, N):
+    #     """
+    #     Recursively, O(n)
+    #     :type N: int
+    #     :rtype: int
+    #     """
+    #     if N == 0:
+    #         return 0
+    #     if N == 1:
+    #         return 1
+    #     return self.fib(N - 1) + self.fib(N - 2)