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★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★You have a number of envelopes with widths and heights given as a pair of integers (w, h)
. One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.html
What is the maximum number of envelopes can you Russian doll? (put one inside other)git
Note:
Rotation is not allowed.github
Example:微信
Input: [[5,4],[6,4],[6,7],[2,3]]
Output: 3 Explanation: The maximum number of envelopes you can Russian doll is ([2,3] => [5,4] => [6,7]).3
给定一些标记了宽度和高度的信封,宽度和高度以整数对形式 (w, h)
出现。当另外一个信封的宽度和高度都比这个信封大的时候,这个信封就能够放进另外一个信封里,如同俄罗斯套娃同样。app
请计算最多能有多少个信封能组成一组“俄罗斯套娃”信封(便可以把一个信封放到另外一个信封里面)。ide
说明:
不容许旋转信封。spa
示例:code
输入: envelopes = 输出: 3 解释: 最多信封的个数为 [2,3] => [5,4] => [6,7]。[[5,4],[6,4],[6,7],[2,3]]3, 组合为:
356ms
1 class Solution { 2 //二分法 3 func maxEnvelopes(_ envelopes: [[Int]]) -> Int { 4 var envelopes = envelopes 5 var dp:[Int] = [Int]() 6 envelopes.sort(by:sortArray) 7 for i in 0..<envelopes.count 8 { 9 var left:Int = 0 10 var right:Int = dp.count 11 var t:Int = envelopes[i][1] 12 while(left < right) 13 { 14 var mid:Int = left + (right - left) / 2 15 if dp[mid] < t 16 { 17 left = mid + 1 18 } 19 else 20 { 21 right = mid 22 } 23 } 24 if right >= dp.count 25 { 26 dp.append(t) 27 } 28 else 29 { 30 dp[right] = t 31 } 32 } 33 return dp.count 34 } 35 36 func sortArray(_ a:[Int],_ b:[Int]) -> Bool 37 { 38 if a[0] == b[0] 39 { 40 return a[1] > b[1] 41 } 42 return a[0] < b[0] 43 } 44 }
6948mshtm
1 class Solution { 2 func maxEnvelopes(_ envelopes: [[Int]]) -> Int { 3 if envelopes.count == 0 { return 0 } 4 let sorted = envelopes.sorted { 5 return $0[0] < $1[0] 6 } 7 8 var dp: [Int] = [Int](repeating: 0, count: envelopes.count) 9 var res = 0 10 for i in 0..<sorted.count { 11 dp[i] = 1 12 for j in 0..<i { 13 // if i is bigger than j 14 if sorted[i][0] > sorted[j][0] && sorted[i][1] > sorted[j][1] { 15 dp[i] = max(dp[i], dp[j] + 1) 16 } 17 } 18 res = max(res, dp[i]) 19 } 20 return res 21 } 22 }