add solution of problem 215: Kth largest element in an array

Author: ppulse

KthLargestElementInAnArray215/README.md

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+Find the **k**th largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element.
+
+For example,
+Given `[3,2,1,5,6,4]` and k = 2, return 5.
+
+#####Note: 
+You may assume k is always valid, 1 ≤ k ≤ array's length.

KthLargestElementInAnArray215/Solution.java

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+public class Solution {
+    public int findKthLargest(int[] nums, int k) {
+        if (nums == null)
+            return 0;
+        
+        int[] minHeap = new int[k + 1];
+        
+        int idxOfNums = 0;
+        int idxOfMinHeap = 1;
+        int numsLen = nums.length;
+        
+        while (idxOfNums < k) {
+            minHeap[idxOfMinHeap++] = nums[idxOfNums++];
+        }
+        
+        buildMinHeap(minHeap, k);
+        
+        while (idxOfNums < numsLen) {
+            if (nums[idxOfNums] > minHeap[1]) {
+                minHeap[1] = nums[idxOfNums];
+                siftDown(minHeap, 1, k);
+            }
+            
+            idxOfNums++;
+        }
+        
+        return minHeap[1];
+    }
+    
+    private void buildMinHeap(int[] minHeap, int heapSize) {
+        int sentinel = heapSize >> 1;
+        for (int i = sentinel; i >= 1; i--)
+            siftDown(minHeap, i, heapSize);
+    }
+    
+    private void siftDown(int[] minHeap, int idx, int heapSize) {
+        int sentinel = heapSize >> 1;
+        
+        while (idx <= sentinel) {
+            int j = idx << 1;
+            if (j + 1 <= heapSize && minHeap[j] > minHeap[j+1])
+                j++;
+            
+            if (minHeap[j] < minHeap[idx]) {
+                int tmp = minHeap[j];
+                minHeap[j] = minHeap[idx];
+                minHeap[idx] = tmp;
+                idx = j;
+            } else {
+                break;
+            }
+        }
+    }
+}