feat: leetcode 56

java8/src/main/java/com/ggggght/learningjava8/leetcode/Leetcode56.java

@@ -0,0 +1,61 @@
+package com.ggggght.learningjava8.leetcode;
+
+import java.util.Collections;
+import java.util.Comparator;
+import java.util.LinkedList;
+import java.util.PriorityQueue;
+
+/**
+ * <pre>
+ *
+ * Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
+ *
+ *
+ *
+ * Example 1:
+ *
+ * Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
+ * Output: [[1,6],[8,10],[15,18]]
+ * Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
+ * Example 2:
+ *
+ * Input: intervals = [[1,4],[4,5]]
+ * Output: [[1,5]]
+ * Explanation: Intervals [1,4] and [4,5] are considered overlapping.
+ *
+ *
+ * Constraints:
+ *
+ * 1 <= intervals.length <= 104
+ * intervals[i].length == 2
+ * 0 <= starti <= endi <= 104
+ * </pre>
+ *
+ */
+public class Leetcode56 {
+    /**
+     * 使用优先队列进行排序对所有的区间进行排序  排序完成之后, 依次遍历第一个元素的第二项, 如果和上一项有重叠, 则更新, 否则就添加当前项
+     * @param intervals
+     * @return
+     */
+    public int[][] merge(int[][] intervals) {
+        PriorityQueue<int[]> q = new PriorityQueue<>(intervals.length, Comparator.comparingInt(o -> o[0]));
+        Collections.addAll(q, intervals);
+        LinkedList<int[]> res = new LinkedList<>();
+        res.add(q.poll());
+        while (!q.isEmpty()) {
+            var poll = q.poll();
+            var last = res.getLast();
+            if (poll[0] <= last[1]) {
+                if (poll[1] > last[1]) {
+                    last[1] = poll[1];
+                    res.set(res.size() - 1, last);
+                }
+            } else {
+                res.add(poll);
+            }
+        }
+
+        return res.toArray(new int[0][0]);
+    }
+}