fd

Author: dingjikerbo

solution/src/main/java/com/inuker/solution/KthLargestElementInArray.java

@@ -12,7 +12,7 @@ public class KthLargestElementInArray {
 
     // 耗时15ms，时间复杂度O(nlgk)，空间复杂度O(k)
     // 按升序排的，出队列k次获取第k大的数
-    public int findKthLargest(int[] nums, int k) {
+    public int findKthLargest2(int[] nums, int k) {
         PriorityQueue<Integer> queue = new PriorityQueue<>();
         for (int n : nums) {
             queue.offer(n);
@@ -34,30 +34,34 @@ public int findKthLargest3(int[] nums, int k) {
     // 对比快速排序T(n) = 2T(n / 2) + n = O(nlgn)
     // 区别在于这个被pivot分隔后，只用处理其中的一半，而快排两边都要处理
     // 耗时18ms
-    public int findKthLargest4(int[] nums, int k) {
-        return nums[findKthLargest4(nums, 0, nums.length - 1, k)];
+    public int findKthLargest(int[] nums, int k) {
+        return findKthLargest(nums, 0, nums.length - 1, k);
     }
 
-    public int findKthLargest4(int[] nums, int start, int end, int k) {
+    public int findKthLargest(int[] nums, int start, int end, int k) {
         int pivot = partition(nums, start, end);
 
         int rank = end - pivot + 1;
 
         if (rank == k) {
-            return pivot;
+            return nums[pivot];
         } else if (rank > k) {
-            return findKthLargest4(nums, pivot + 1, end, k);
+            return findKthLargest(nums, pivot + 1, end, k);
         } else {
-            return findKthLargest4(nums, start, pivot - 1, k - rank);
+            return findKthLargest(nums, start, pivot - 1, k - rank);
         }
     }
 
     public int partition(int[] nums, int start, int end) {
-        int pivot = nums[end], left = start;
+        int pivot = nums[end], left = start, right = end - 1;
 
-        for (int i = start; i < end; i++) {
-            if (nums[i] <= pivot) {
-                swap(nums, left++, i);
+        for (int i = start; i <= right; ) {
+            if (nums[i] < pivot) {
+                swap(nums, left++, i++);
+            } else if (nums[i] > pivot) {
+                swap(nums, right--, i);
+            } else {
+                i++;
             }
         }
 

solution/src/main/java/com/inuker/solution/MedianOfTwoSortedArrays.java

@@ -2,6 +2,7 @@
 
 /**
  * Created by dingjikerbo on 16/12/20.
+ * https://leetcode.com/articles/median-of-two-sorted-arrays/
  */
 
 public class MedianOfTwoSortedArrays {

solution/src/main/java/com/inuker/solution/MergeKSortedList.java

@@ -8,6 +8,7 @@
 
 /**
  * Created by dingjikerbo on 16/11/20.
+ * https://leetcode.com/articles/merge-k-sorted-list/
  */
 
 public class MergeKSortedList {
@@ -23,13 +24,7 @@ public ListNode mergeKLists(ListNode[] lists) {
         PriorityQueue<ListNode> queue = new PriorityQueue<>(new Comparator<ListNode>() {
             @Override
             public int compare(ListNode node1, ListNode node2) {
-                if (node1.val == node2.val) {
-                    return 0;
-                } else if (node1.val < node2.val) {
-                    return -1;
-                } else {
-                    return 1;
-                }
+                return node1.val - node2.val;
             }
         });
 

solution/src/main/java/com/inuker/solution/WiggleSortII.java

@@ -34,6 +34,7 @@ public void wiggleSort2(int[] nums) {
      * 只要这样交叉放着就能符合wiggle
      * 直观的做法是另外开辟一个空间来放wiggle结果
      * 如果要求不开辟空间，则只能用坐标映射了
+     * 类似的可以参考sort colors和find kth largest element
      */
     // 时间复杂度O(n)，空间复杂度O(l)
     public void wiggleSort(int[] nums) {

test/src/main/java/com/inuker/test/main.java

@@ -26,18 +26,10 @@ public static void main(String[] args) {
         int[] arr = new int[] {
                 6, 13, 5, 4, 5, 2
         };
-        new WiggleSortII().wiggleSort(arr);
+        int ff = new fd().findKthLargest(arr, 6);
+        System.out.println(ff);
         for (int n : arr) {
             System.out.print(n + " ");
         }
     }
-
-    public void wiggleSort2(int[] nums) {
-        int[] arr = nums.clone();
-        Arrays.sort(arr);
-        int n = nums.length, j = (n - 1) / 2, k = n - 1;
-        for (int i = 0; i < nums.length; i++) {
-            nums[i] = i % 2 == 0 ? arr[j--] : arr[k--];
-        }
-    }
 }