Create Majorityelement#229

Medium/Readme.md

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+Problem #229 (Majority Element II | Array, Counting, Hash Table, Sorting)
+Medium
+
+Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
+
+Example 1
+Input: nums = [3,2,3] **Output:** [3]`
+
+Example 2:
+Input: nums = [1]
+Output: [1]
+
+Example 3:
+Input: nums = [1,2]
+Output: [1,2]
+
+Constraints:
+. 1<=nums.length<= 5*104
+. -109<=nums[i]<=10^9

Medium/solution.cpp

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+class Solution {
+ public:
+  vector<int> majorityElement(vector<int>& nums) {
+    vector<int> ans;
+    int candidate1 = 0;
+    int candidate2 = 1;   // Any number different from candidate1
+    int countSoFar1 = 0;  // # of candidate1 so far
+    int countSoFar2 = 0;  // # of candidate2 so far
+
+    for (const int num : nums)
+      if (num == candidate1) {
+        ++countSoFar1;
+      } else if (num == candidate2) {
+        ++countSoFar2;
+      } else if (countSoFar1 == 0) {  // Assign new candidate
+        candidate1 = num;
+        ++countSoFar1;
+      } else if (countSoFar2 == 0) {  // Assign new candidate
+        candidate2 = num;
+        ++countSoFar2;
+      } else {  // Meet a new number, so pair out previous counts
+        --countSoFar1;
+        --countSoFar2;
+      }
+
+    const int count1 = count(nums.begin(), nums.end(), candidate1);
+    const int count2 = count(nums.begin(), nums.end(), candidate2);
+
+    if (count1 > nums.size() / 3)
+      ans.push_back(candidate1);
+    if (count2 > nums.size() / 3)
+      ans.push_back(candidate2);
+    return ans;
+  }
+};

Medium/solution.java

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+class Solution {
+  public List<Integer> majorityElement(int[] nums) {
+    List<Integer> ans = new ArrayList<>();
+    int candidate1 = 0;
+    int candidate2 = 1;  // Any number different from candidate1
+    int countSoFar1 = 0; // # of candidate1 so far
+    int countSoFar2 = 0; // # of candidate2 so far
+
+    for (final int num : nums)
+      if (num == candidate1) {
+        ++countSoFar1;
+      } else if (num == candidate2) {
+        ++countSoFar2;
+      } else if (countSoFar1 == 0) { // Assign new candidate
+        candidate1 = num;
+        ++countSoFar1;
+      } else if (countSoFar2 == 0) { // Assign new candidate
+        candidate2 = num;
+        ++countSoFar2;
+      } else { // Meet a new number, so pair out previous counts
+        --countSoFar1;
+        --countSoFar2;
+      }
+
+    int count1 = 0;
+    int count2 = 0;
+
+    for (final int num : nums)
+      if (num == candidate1)
+        ++count1;
+      else if (num == candidate2)
+        ++count2;
+
+    if (count1 > nums.length / 3)
+      ans.add(candidate1);
+    if (count2 > nums.length / 3)
+      ans.add(candidate2);
+    return ans;
+  }
+}