feat(leecode): 56.合并区间

docs/10.LeeCode/56.合并区间.md

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+---
+title: 56.合并区间
+date: 2024-03-24
+lang: 'zh-CN'
+sidebar: 'auto'
+categories:
+  - LeeCode
+tags:
+location: HangZhou
+---
+
+# Heading
+
+[[toc]]
+
+[56.合并区间](https://leetcode.cn/problems/merge-intervals/description/)
+
+Tags: algorithms bloomberg facebook google linkedin microsoft twitter yelp array sort
+
+Langs: c cpp csharp dart elixir erlang golang java javascript kotlin php python python3 racket ruby rust scala swift typescript
+
+- algorithms
+- Medium (49.82%)
+- Likes: 2274
+- Dislikes: -
+- Total Accepted: 821.3K
+- Total Submissions: 1.6M
+- Testcase Example: '[[1,3],[2,6],[8,10],[15,18]]'
+
+<p>以数组 <code>intervals</code> 表示若干个区间的集合，其中单个区间为 <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> 。请你合并所有重叠的区间，并返回&nbsp;<em>一个不重叠的区间数组，该数组需恰好覆盖输入中的所有区间</em>&nbsp;。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>intervals = [[1,3],[2,6],[8,10],[15,18]]
+<strong>输出：</strong>[[1,6],[8,10],[15,18]]
+<strong>解释：</strong>区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
+</pre>
+
+<p><strong>示例&nbsp;2：</strong></p>
+
+<pre>
+<strong>输入：</strong>intervals = [[1,4],[4,5]]
+<strong>输出：</strong>[[1,5]]
+<strong>解释：</strong>区间 [1,4] 和 [4,5] 可被视为重叠区间。</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= intervals.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>intervals[i].length == 2</code></li>
+	<li><code>0 &lt;= start<sub>i</sub> &lt;= end<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
+</ul>
+
+<<< @/src/LeeCode/56.合并区间.js

src/LeeCode/56.合并区间.js

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+/*
+ * @lc app=leetcode.cn id=56 lang=javascript
+ *
+ * [56] 合并区间
+ */
+
+// @lc code=start
+/**
+ * @param {number[][]} intervals
+ * @return {number[][]}
+ */
+var merge = function (intervals) {
+  if (intervals.length <= 1) {
+    return intervals
+  }
+
+  let insertion
+  for (let i = 1; i < intervals.length; i++) {
+    insertion = intervals[i]
+    let j = i
+    while (j > 0 && insertion[0] < intervals[j - 1][0]) {
+      intervals[j] = intervals[j - 1]
+      j--
+    }
+    intervals[j] = insertion
+  }
+
+  const result = [intervals[0]]
+  for (let i = 1; i < intervals.length; i++) {
+    if (result[result.length - 1][1] >= intervals[i][0]) {
+      result[result.length - 1][1] = Math.max(result[result.length - 1][1], intervals[i][1])
+    } else {
+      result.push(intervals[i])
+    }
+  }
+  return result
+}
+// @lc code=end