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➤微信公众号：山青咏芝（shanqingyongzhi）➤博客园地址：山青咏芝（）➤GitHub地址：➤原文地址： ➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★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: 2Explanation: The longest chain is [1,2] -> [3,4]
Note:
- 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：
- 给出数对的个数在 [1, 1000] 范围内。
Runtime: 236 ms
Memory Usage: 19.1 MB
1 class Solution { 2 func findLongestChain(_ pairs: [[Int]]) -> Int { 3 var pairs = pairs 4 var res:Int = 0 5 var end:Int = Int.min 6 pairs.sort(by:{(a:[Int],b:[Int]) -> Bool in 7 return a[1] < b[1] 8 }) 9 for pair in pairs10 {11 if pair[0] > end12 {13 res += 114 end = pair[1]15 }16 }17 return res18 }19 } 284ms
1 class Solution { 2 func findLongestChain(_ pairs: [[Int]]) -> Int { 3 let sortPairs = pairs.sorted { $0[1] < $1[1] } 4 var ans = 0, curEnd = Int.min 5 for pair in sortPairs { 6 if pair[0] > curEnd { 7 ans += 1 8 curEnd = pair[1] 9 }10 }11 return ans12 }13 } 296ms
1 class Solution { 2 func findLongestChain(_ pairs: [[Int]]) -> Int { 3 guard pairs.count > 1 else{ 4 return pairs.count 5 } 6 let sorted = pairs.sorted(by: {$0[1] < $1[1]}) 7 var count : Int = 1 8 var currentPair = sorted[0] 9 for pair in sorted{10 if pair.first! > currentPair.last!{11 count += 112 currentPair = pair13 }14 } 15 return count 16 }17 } 1260ms
1 sample 1260 ms submission 2 class Solution { 3 func findLongestChain(_ pairs: [[Int]]) -> Int { 4 let n = pairs.count 5 if n <= 1{ return n } 6 7 let sorted = pairs.sorted { $0[0] < $1[0] } 8 var f = [Int](repeating: 1, count: n) 9 var result = 110 11 for i in 1.. 1272ms
1 class Solution { 2 func findLongestChain(_ pairs: [[Int]]) -> Int { 3 guard pairs.count > 1 else { return 1 } 4 let pairs = pairs.sorted(by: { 5 return $0[0] < $1[0] 6 }) 7 8 let n = pairs.count 9 var dp: [Int] = Array(repeating: 0, count: n)10 dp[0] = 111 12 for i in 1 ..< n {13 for j in 0 ..< i {14 if pairs[i][0] > pairs[j][1] {15 dp[i] = max(dp[i], dp[j])16 }17 }18 dp[i] += 119 }20 21 return dp[n - 1]22 }23 }