961_N-Repeated_Element_in_Size_2N_Array

Author: qiyuangong

README.md

@@ -194,6 +194,7 @@ Also, there are open source implementations for basic data structs and algorithm
 | 946 | [Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/) | [Python](https://github.com/qiyuangong/leetcode/blob/master/python/946_Validate_Stack_Sequences.py) [Java](https://github.com/qiyuangong/leetcode/blob/master/java/946_Validate_Stack_Sequences.java) | Add a stack named inStack to help going through pushed and popped. O(n) and O(n) |
 | 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) |
 
 | # | To Understand |
 |---| ----- |

java/961_N-Repeated_Element_in_Size_2N_Array.java

@@ -0,0 +1,12 @@
+class Solution {
+    public int repeatedNTimes(int[] A) {
+        HashMap<Integer, Integer> hash = new HashMap<>();
+        int ans = A[0];
+        for (int n: A) {
+            int count = hash.getOrDefault(n, 0) + 1;
+            hash.put(n, count);
+            if (count >= hash.get(ans)) ans = n;
+        }
+        return ans;
+    }
+}

python/961_N-Repeated_Element_in_Size_2N_Array.py

@@ -0,0 +1,18 @@
+import collections
+
+
+class Solution(object):
+    def repeatedNTimes(self, A):
+        """
+        :type A: List[int]
+        :rtype: int
+        """
+        counter = collections.Counter(A)
+        return counter.most_common(1)[0][0]
+
+
+if __name__ == '__main__':
+    s = Solution()
+    print s.repeatedNTimes([1, 2, 3, 3])
+    print s.repeatedNTimes([2, 1, 2, 5, 3, 2])
+    print s.repeatedNTimes([5, 1, 5, 2, 5, 3, 5, 4])