[Swift]LeetCode757. 设置交集大小至少为2 | Set Intersection Size At Least Two

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An integer interval [a, b] (for integers a < b) is a set of all consecutive integers from ato b, including a and b.

Find the minimum size of a set S such that for every integer interval A in intervals, the intersection of S with A has size at least 2.

Example 1:

Input: intervals = [[1, 3], [1, 4], [2, 5], [3, 5]]
Output: 3
Explanation:
Consider the set S = {2, 3, 4}.  For each interval, there are at least 2 elements from S in the interval.
Also, there isn't a smaller size set that fulfills the above condition.
Thus, we output the size of this set, which is 3. 

Example 2:

Input: intervals = [[1, 2], [2, 3], [2, 4], [4, 5]]
Output: 5
Explanation:
An example of a minimum sized set is {1, 2, 3, 4, 5}. 

Note:

1. intervals will have length in range [1, 3000].
2. intervals[i] will have length 2, representing some integer interval.
3. intervals[i][j] will be an integer in [0, 10^8].

---

一个整数区间 [a, b]  ( a < b ) 代表着从 a 到 b 的所有连续整数，包括 a 和 b。

给你一组整数区间intervals，请找到一个最小的集合 S，使得 S 里的元素与区间intervals中的每一个整数区间都至少有2个元素相交。

输出这个最小集合S的大小。

示例 1:

输入: intervals = [[1, 3], [1, 4], [2, 5], [3, 5]]
输出: 3
解释:
考虑集合 S = {2, 3, 4}. S与intervals中的四个区间都有至少2个相交的元素。
且这是S最小的情况，故我们输出3。

示例 2:

输入: intervals = [[1, 2], [2, 3], [2, 4], [4, 5]]
输出: 5
解释:
最小的集合S = {1, 2, 3, 4, 5}.

注意:

1. intervals 的长度范围为[1, 3000]。
2. intervals[i] 长度为 2，分别代表左、右边界。
3. intervals[i][j] 的值是 [0, 10^8]范围内的整数。

---

Runtime: 252 ms

Memory Usage: 19.1 MB

class Solution {
    func intersectionSizeTwo(_ intervals: [[Int]]) -> Int {
        var intervals = intervals
        var res:Int = 0
        var p1:Int = -1
        var p2:Int = -1
        intervals.sort(by:{(a:[Int],b:[Int]) -> Bool in 
                          return a[1] < b[1] || (a[1] == b[1] && a[0] > b[0])})
        for interval in intervals
        {
            if interval[0] <= p1 {continue}
            if interval[0] > p2
            {
                res += 2
                p2 = interval[1]
                p1 = p2 - 1
            }
            else
            {
                res += 1
                p1 = p2
                p2 = interval[1]
            }
        }
        return res       
    }
}