From a9be61f016e1816c097d467ffd7a1a1ae5a4f232 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Mon, 6 Sep 2021 13:07:58 +0800
Subject: [PATCH] =?UTF-8?q?435:=E6=97=A0=E9=87=8D=E5=8F=A0=E5=8C=BA?=
 =?UTF-8?q?=E9=97=B4?=
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---
 .../editor/cn/NonOverlappingIntervals.java    | 74 +++++++++++++++++++
 .../cn/doc/content/NonOverlappingIntervals.md | 39 ++++++++++
 2 files changed, 113 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/NonOverlappingIntervals.java
 create mode 100644 src/main/java/leetcode/editor/cn/doc/content/NonOverlappingIntervals.md

diff --git a/src/main/java/leetcode/editor/cn/NonOverlappingIntervals.java b/src/main/java/leetcode/editor/cn/NonOverlappingIntervals.java
new file mode 100644
index 0000000..5afca07
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/NonOverlappingIntervals.java
@@ -0,0 +1,74 @@
+//给定一个区间的集合，找到需要移除区间的最小数量，使剩余区间互不重叠。 
+//
+// 注意: 
+//
+// 
+// 可以认为区间的终点总是大于它的起点。 
+// 区间 [1,2] 和 [2,3] 的边界相互“接触”，但没有相互重叠。 
+// 
+//
+// 示例 1: 
+//
+// 
+//输入: [ [1,2], [2,3], [3,4], [1,3] ]
+//
+//输出: 1
+//
+//解释: 移除 [1,3] 后，剩下的区间没有重叠。
+// 
+//
+// 示例 2: 
+//
+// 
+//输入: [ [1,2], [1,2], [1,2] ]
+//
+//输出: 2
+//
+//解释: 你需要移除两个 [1,2] 来使剩下的区间没有重叠。
+// 
+//
+// 示例 3: 
+//
+// 
+//输入: [ [1,2], [2,3] ]
+//
+//输出: 0
+//
+//解释: 你不需要移除任何区间，因为它们已经是无重叠的了。
+// 
+// Related Topics 贪心 数组 动态规划 排序 👍 491 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.Arrays;
+import java.util.Comparator;
+
+//435:无重叠区间
+class NonOverlappingIntervals {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new NonOverlappingIntervals().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int eraseOverlapIntervals(int[][] intervals) {
+            if (intervals.length == 0) {
+                return 0;
+            }
+            Arrays.sort(intervals, Comparator.comparingInt(interval -> interval[1]));
+            int num = intervals[0][1];
+            int count = 1;
+            for (int i = 1; i < intervals.length; ++i) {
+                if (intervals[i][0] >= num) {
+                    ++count;
+                    num = intervals[i][1];
+                }
+            }
+            return intervals.length - count;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/doc/content/NonOverlappingIntervals.md b/src/main/java/leetcode/editor/cn/doc/content/NonOverlappingIntervals.md
new file mode 100644
index 0000000..0a37277
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+++ b/src/main/java/leetcode/editor/cn/doc/content/NonOverlappingIntervals.md
@@ -0,0 +1,39 @@
+<p>给定一个区间的集合，找到需要移除区间的最小数量，使剩余区间互不重叠。</p>
+
+<p><strong>注意:</strong></p>
+
+<ol>
+	<li>可以认为区间的终点总是大于它的起点。</li>
+	<li>区间 [1,2] 和 [2,3] 的边界相互&ldquo;接触&rdquo;，但没有相互重叠。</li>
+</ol>
+
+<p><strong>示例 1:</strong></p>
+
+<pre>
+<strong>输入:</strong> [ [1,2], [2,3], [3,4], [1,3] ]
+
+<strong>输出:</strong> 1
+
+<strong>解释:</strong> 移除 [1,3] 后，剩下的区间没有重叠。
+</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre>
+<strong>输入:</strong> [ [1,2], [1,2], [1,2] ]
+
+<strong>输出:</strong> 2
+
+<strong>解释:</strong> 你需要移除两个 [1,2] 来使剩下的区间没有重叠。
+</pre>
+
+<p><strong>示例 3:</strong></p>
+
+<pre>
+<strong>输入:</strong> [ [1,2], [2,3] ]
+
+<strong>输出:</strong> 0
+
+<strong>解释:</strong> 你不需要移除任何区间，因为它们已经是无重叠的了。
+</pre>
+<div><div>Related Topics</div><div><li>贪心</li><li>数组</li><li>动态规划</li><li>排序</li></div></div><br><div><li>👍 491</li><li>👎 0</li></div>
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