adding the heapify solution

leetcode/p912_heapsort_array_heapify.cpp

@@ -0,0 +1,46 @@
+class Solution {
+public:
+    void heapify(vector<int>& nums, int n, int i){
+        int largest = i; // assuming i is the largest
+        int left = 2*i+1;    // i's left child is at this location
+        int right = 2*i+2;   // i's right child is at this location
+
+        if(left<n && nums[left]>nums[largest]){
+            largest = left;
+        }
+
+        if(right<n && nums[right]>nums[largest]){
+            largest = right;
+        }
+
+        // after the above, largest basically will either stay the same, or will be either left or right, depending on nums[left] is larger or nums[right] is larger. largest stays the same if it is already larger than its two children.
+        // if largest is changed, then we do need to swap.
+        if(largest != i){
+            std::swap(nums[i], nums[largest]);
+            heapify(nums, n, largest);
+        }
+    }
+
+    // heap sort: O(nlogn)
+    vector<int> sortArray(vector<int>& nums) {
+        int n = nums.size();
+        // build the heap, starting from the last non-leaf node
+        for(int i=n/2-1; i>=0; i--){
+            // heapify the subtree whose root is at i
+            // i.e., build a max heap, with i being the root.
+            heapify(nums, n, i);
+        }
+
+        // now the first one is the largest, swap it to the back
+        // do this n-1 times.
+        for(int i=0; i<(n-1); i++){
+            // nums[0] is always the largest one
+            std::swap(nums[0], nums[n-1-i]);
+            // build the max heap again, with 0 being the root.
+            // but only consider n-1-i elements, as the others are already in the right place.
+            heapify(nums, n-1-i, 0);
+        }
+
+        return nums;
+    }
+};