[Swift]LeetCode646. 最长数对链 | Maximum Length of Pair Chain

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You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4] 

Note:

1. The number of given pairs will be in the range [1, 1000].

---

给出 n 个数对。 在每一个数对中，第一个数字总是比第二个数字小。

现在，我们定义一种跟随关系，当且仅当 b < c 时，数对(c, d) 才可以跟在 (a, b) 后面。我们用这种形式来构造一个数对链。

给定一个对数集合，找出能够形成的最长数对链的长度。你不需要用到所有的数对，你可以以任何顺序选择其中的一些数对来构造。

示例 :

输入: [[1,2], [2,3], [3,4]]
输出: 2
解释: 最长的数对链是 [1,2] -> [3,4]

注意：

1. 给出数对的个数在 [1, 1000] 范围内。

---

Runtime: 236 ms

Memory Usage: 19.1 MB

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        var pairs = pairs
        var res:Int = 0
        var end:Int = Int.min
        pairs.sort(by:{(a:[Int],b:[Int]) -> Bool in 
                       return a[1] < b[1]                      
                      })
        for pair in pairs
        {
            if pair[0] > end
            {
                res += 1
                end = pair[1]
            }
        }
        return res
    }
}

---

284ms

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        let sortPairs = pairs.sorted { $0[1] < $1[1] }
        var ans = 0, curEnd = Int.min
        for pair in sortPairs {
            if pair[0] > curEnd {
                ans += 1
                curEnd = pair[1]
            }
        }
        return ans
    }
}

---

296ms

class Solution {   
    func findLongestChain(_ pairs: [[Int]]) -> Int {
    guard pairs.count > 1 else{
        return pairs.count
    }
    let sorted = pairs.sorted(by: {$0[1] < $1[1]})
    var count : Int = 1
    var currentPair = sorted[0]
    for pair in sorted{
        if pair.first! > currentPair.last!{
            count += 1
            currentPair = pair
        }
    }    
    return count    
    }
}

---

1260ms

sample 1260 ms submission
class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        let n = pairs.count
        if n <= 1{ return n }
        
        let sorted = pairs.sorted { $0[0] < $1[0] }
        var f = [Int](repeating: 1, count: n)
        var result = 1
        
        for i in 1..<n {
            for j in 0..<i {
                if sorted[j][1] < sorted[i][0] {
                    f[i] = max(f[i], f[j] + 1)
                    result = max(result, f[i])
                }
            }
        }
        
        return result
    }
}

---

1272ms

class Solution {
    func findLongestChain(_ pairs: [[Int]]) -> Int {
        guard pairs.count > 1 else { return 1 }
        let pairs = pairs.sorted(by: {
            return $0[0] < $1[0]
        })
        
        let n = pairs.count
        var dp: [Int] = Array(repeating: 0, count: n)
        dp[0] = 1
        
        for i in  1 ..< n {
            for j in 0 ..< i {
                if pairs[i][0] > pairs[j][1] {
                    dp[i] = max(dp[i], dp[j])
                }
            }
            dp[i] += 1
        }
        
        return dp[n - 1]
    }
}