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Description
tab: English
<p>You are given an array of non-overlapping intervals <code>intervals</code> where <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> represent the start and the end of the <code>i<sup>th</sup></code> interval and <code>intervals</code> is sorted in ascending order by <code>start<sub>i</sub></code>. You are also given an interval <code>newInterval = [start, end]</code> that represents the start and end of another interval.</p>
<p>Insert <code>newInterval</code> into <code>intervals</code> such that <code>intervals</code> is still sorted in ascending order by <code>start<sub>i</sub></code> and <code>intervals</code> still does not have any overlapping intervals (merge overlapping intervals if necessary).</p>
<p>Return <code>intervals</code><em> after the insertion</em>.</p>
<p><strong>Note</strong> that you don't need to modify <code>intervals</code> in-place. You can make a new array and return it.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> intervals = [[1,3],[6,9]], newInterval = [2,5]
<strong>Output:</strong> [[1,5],[6,9]]
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
<strong>Output:</strong> [[1,2],[3,10],[12,16]]
<strong>Explanation:</strong> Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>0 <= intervals.length <= 10<sup>4</sup></code></li>
<li><code>intervals[i].length == 2</code></li>
<li><code>0 <= start<sub>i</sub> <= end<sub>i</sub> <= 10<sup>5</sup></code></li>
<li><code>intervals</code> is sorted by <code>start<sub>i</sub></code> in <strong>ascending</strong> order.</li>
<li><code>newInterval.length == 2</code></li>
<li><code>0 <= start <= end <= 10<sup>5</sup></code></li>
</ul>
> [!tip]- Hint 1
>
> Intervals Array is sorted. Can you use Binary Search to find the correct position to insert the new Interval.?
> [!tip]- Hint 2
>
> Can you try merging the overlapping intervals while inserting the new interval?
> [!tip]- Hint 3
>
> This can be done by comparing the end of the last interval with the start of the new interval and vice versa.
---
[submissions](https://leetcode.com/problems/insert-interval/submissions/) | [solutions](https://leetcode.com/problems/insert-interval/solutions/)
tab: 中文
<p>给你一个<strong> 无重叠的</strong><em> ，</em>按照区间起始端点排序的区间列表 <code>intervals</code>，其中 <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> 表示第 <code>i</code> 个区间的开始和结束，并且 <code>intervals</code> 按照 <code>start<sub>i</sub></code> 升序排列。同样给定一个区间 <code>newInterval = [start, end]</code> 表示另一个区间的开始和结束。</p>
<p>在 <code>intervals</code> 中插入区间 <code>newInterval</code>，使得 <code>intervals</code> 依然按照 <code>start<sub>i</sub></code> 升序排列，且区间之间不重叠（如果有必要的话，可以合并区间）。</p>
<p>返回插入之后的 <code>intervals</code>。</p>
<p><strong>注意</strong> 你不需要原地修改 <code>intervals</code>。你可以创建一个新数组然后返回它。</p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>intervals = [[1,3],[6,9]], newInterval = [2,5]
<strong>输出：</strong>[[1,5],[6,9]]
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
<strong>输出：</strong>[[1,2],[3,10],[12,16]]
<strong>解释：</strong>这是因为新的区间 <code>[4,8]</code> 与 <code>[3,5],[6,7],[8,10]</code> 重叠。
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>0 <= intervals.length <= 10<sup>4</sup></code></li>
<li><code>intervals[i].length == 2</code></li>
<li><code>0 <= start<sub>i</sub> <= end<sub>i</sub> <= 10<sup>5</sup></code></li>
<li><code>intervals</code> 根据 <code>start<sub>i</sub></code> 按 <strong>升序</strong> 排列</li>
<li><code>newInterval.length == 2</code></li>
<li><code>0 <= start <= end <= 10<sup>5</sup></code></li>
</ul>
> [!tip]- 提示 1
>
> Intervals Array is sorted. Can you use Binary Search to find the correct position to insert the new Interval.?
> [!tip]- 提示 2
>
> Can you try merging the overlapping intervals while inserting the new interval?
> [!tip]- 提示 3
>
> This can be done by comparing the end of the last interval with the start of the new interval and vice versa.
---
[提交记录](https://leetcode.cn/problems/insert-interval/submissions/) | [题解](https://leetcode.cn/problems/insert-interval/solution/)
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