[Swift]LeetCode352. 将数据流变为多个不相交间隔 | Data Stream as Disjoint Intervals

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Given a data stream input of non-negative integers a1, a2, ..., an, ..., summarize the numbers seen so far as a list of disjoint intervals.

For example, suppose the integers from the data stream are 1, 3, 7, 2, 6, ..., then the summary will be:

[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7]

Follow up:
What if there are lots of merges and the number of disjoint intervals are small compared to the data stream's size?

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给定一个非负整数的数据流输入 a1，a2，…，an，…，将到目前为止看到的数字总结为不相交的间隔列表。

例如，假设数据流中的整数为 1，3，7，2，6，…，每次的总结为：

[1, 1]
[1, 1], [3, 3]
[1, 1], [3, 3], [7, 7]
[1, 3], [7, 7]
[1, 3], [6, 7] 

进阶：
如果有很多合并，并且与数据流的大小相比，不相交间隔的数量很小，该怎么办?

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412ms

/**
 * Definition for an interval.
 * public class Interval {
 *   public var start: Int
 *   public var end: Int
 *   public init(_ start: Int, _ end: Int) {
 *     self.start = start
 *     self.end = end
 *   }
 * }
 */

class SummaryRanges {
    var v:[Interval]
    /** Initialize your data structure here. */
    init() {
        v = [Interval]()
    }

    func addNum(_ val: Int) {
        var cur:Interval = Interval(val, val)
        var res:[Interval] = [Interval]()
        var pos:Int = 0
        for a in v
        {
            if cur.end + 1 < a.start
            {
                res.append(a)
            }
            else if cur.start > a.end + 1
            {
                res.append(a)
                pos += 1
            }
            else
            {
                cur.start = min(cur.start, a.start)
                cur.end = max(cur.end, a.end)
            }            
        }
        res.insert( cur,at: pos)
        v = res        
    }

    func getIntervals() -> [Interval] {
        return v
    }
}

/**
 * Your SummaryRanges object will be instantiated and called as such:
 * let obj = SummaryRanges()
 * obj.addNum(val)
 * let ret_2: [Interval] = obj.getIntervals()
 */