solution.md

# 将数据流变为多个不相交区间

<p>&nbsp;给你一个由非负整数&nbsp;<code>a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub></code> 组成的数据流输入，请你将到目前为止看到的数字总结为不相交的区间列表。</p>

<p>实现 <code>SummaryRanges</code> 类：</p>

<div class="original__bRMd">
<div>
<ul>
	<li><code>SummaryRanges()</code> 使用一个空数据流初始化对象。</li>
	<li><code>void addNum(int val)</code> 向数据流中加入整数 <code>val</code> 。</li>
	<li><code>int[][] getIntervals()</code> 以不相交区间&nbsp;<code>[start<sub>i</sub>, end<sub>i</sub>]</code> 的列表形式返回对数据流中整数的总结。</li>
</ul>

<p>&nbsp;</p>

<p><strong>示例：</strong></p>

<pre>
<strong>输入：</strong>
["SummaryRanges", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals", "addNum", "getIntervals"]
[[], [1], [], [3], [], [7], [], [2], [], [6], []]
<strong>输出：</strong>
[null, null, [[1, 1]], null, [[1, 1], [3, 3]], null, [[1, 1], [3, 3], [7, 7]], null, [[1, 3], [7, 7]], null, [[1, 3], [6, 7]]]

<strong>解释：</strong>
SummaryRanges summaryRanges = new SummaryRanges();
summaryRanges.addNum(1);      // arr = [1]
summaryRanges.getIntervals(); // 返回 [[1, 1]]
summaryRanges.addNum(3);      // arr = [1, 3]
summaryRanges.getIntervals(); // 返回 [[1, 1], [3, 3]]
summaryRanges.addNum(7);      // arr = [1, 3, 7]
summaryRanges.getIntervals(); // 返回 [[1, 1], [3, 3], [7, 7]]
summaryRanges.addNum(2);      // arr = [1, 2, 3, 7]
summaryRanges.getIntervals(); // 返回 [[1, 3], [7, 7]]
summaryRanges.addNum(6);      // arr = [1, 2, 3, 6, 7]
summaryRanges.getIntervals(); // 返回 [[1, 3], [6, 7]]
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>0 &lt;= val &lt;= 10<sup>4</sup></code></li>
	<li>最多调用&nbsp;<code>addNum</code> 和 <code>getIntervals</code> 方法 <code>3 * 10<sup>4</sup></code> 次</li>
</ul>
</div>
</div>

<p>&nbsp;</p>

<p><strong>进阶：</strong>如果存在大量合并，并且与数据流的大小相比，不相交区间的数量很小，该怎么办?</p>


## template

```python

class SummaryRanges:
    def __init__(self):
        self.rec = set()

        self.s = list()

    def binarySearch_l(self, x: int):
        s = self.s
        l, r = 0, len(s) - 1
        while l < r:
            mid = l + r >> 1
            if s[mid][0] >= x:
                r = mid
            else:
                l = mid + 1
        return l

    def binarySearch(self, x: int):
        s = self.s
        l, r = 0, len(s) - 1
        while l < r:
            mid = l + r >> 1
            if s[mid][1] >= x:
                r = mid
            else:
                l = mid + 1
        return l

    def addNum(self, val: int) -> None:
        rec, s = self.rec, self.s
        if val in rec:
            return
        else:

            if val - 1 not in rec and val + 1 not in rec:

                s.append([val, val])
                s.sort()
            elif val - 1 in rec and val + 1 not in rec:

                p = self.binarySearch(val - 1)
                s[p][1] = val
            elif val - 1 not in rec and val + 1 in rec:

                p = self.binarySearch_l(val + 1)
                s[p][0] = val
            else:

                p = self.binarySearch(val)
                s[p - 1][1] = s[p][1]

                s.pop(p)
            rec.add(val)

    def getIntervals(self) -> List[List[int]]:
        return self.s

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```