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add leetcode problem-cn part5

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2022-03-27 20:52:13 +08:00
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算法题（国内版）/problem (Chinese)/无重叠区间 [non-overlapping-intervals].html Normal file
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<p>给定一个区间的集合&nbsp;<code>intervals</code>&nbsp;，其中 <code>intervals[i] = [start<sub>i</sub>, end<sub>i</sub>]</code>&nbsp;。返回 <em>需要移除区间的最小数量，使剩余区间互不重叠&nbsp;</em>。</p>
<p>&nbsp;</p>
<p><strong>示例 1:</strong></p>
<pre>
<strong>输入:</strong> intervals = [[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> intervals = [ [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> intervals = [ [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>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= intervals.length &lt;= 10<sup>5</sup></code></li>
<li><code>intervals[i].length == 2</code></li>
<li><code>-5 * 10<sup>4</sup>&nbsp;&lt;= start<sub>i</sub>&nbsp;&lt; end<sub>i</sub>&nbsp;&lt;= 5 * 10<sup>4</sup></code></li>
</ul>