[Swift]LeetCode718. 最长重复子数组 | Maximum Length of Repeated Subarray
作者:互联网
Given two integer arrays A
and B
, return the maximum length of an subarray that appears in both arrays.
Example 1:
Input: A: [1,2,3,2,1] B: [3,2,1,4,7] Output: 3 Explanation: The repeated subarray with maximum length is [3, 2, 1].
Note:
- 1 <= len(A), len(B) <= 1000
- 0 <= A[i], B[i] < 100
给两个整数数组 A
和 B
,返回两个数组中公共的、长度最长的子数组的长度。
示例 1:
输入: A: [1,2,3,2,1] B: [3,2,1,4,7] 输出: 3 解释: 长度最长的公共子数组是 [3, 2, 1]。
说明:
- 1 <= len(A), len(B) <= 1000
- 0 <= A[i], B[i] < 100
308ms
1 class Solution { 2 func findLength(_ A: [Int], _ B: [Int]) -> Int { 3 var ret = 0 4 let lenA = A.count, lenB = B.count 5 for k in 1..<(lenA + lenB) { 6 var i = max(0, lenA - k) 7 var j = max(0, k - lenA) 8 var len = 0 9 while i < lenA && j < lenB { 10 len = (A[i] == B[j]) ? (len + 1) : 0 11 ret = max(ret, len) 12 i += 1 13 j += 1 14 } 15 } 16 return ret 17 } 18 }
1304ms
1 class Solution { 2 func findLength(_ A: [Int], _ B: [Int]) -> Int { 3 let m = A.count, n = B.count 4 var res = 0 5 6 var dp = [[Int]](repeating:[Int](repeating: 0, count: n+1), count: m + 1) 7 8 for i in 1...m { 9 for j in 1...n { 10 if A[i-1] == B[j-1] { 11 dp[i][j] = 1 + dp[i-1][j-1] 12 res = max(res, dp[i][j]) 13 } 14 } 15 } 16 return res 17 } 18 }
Runtime: 2028 ms Memory Usage: 25.4 MB
1 class Solution { 2 func findLength(_ A: [Int], _ B: [Int]) -> Int { 3 var res:Int = 0 4 var dp:[[Int]] = [[Int]](repeating:[Int](repeating:0,count:B.count + 1),count:A.count + 1) 5 for i in 1..<dp.count 6 { 7 for j in 1..<dp[i].count 8 { 9 dp[i][j] = (A[i - 1] == B[j - 1]) ? dp[i - 1][j - 1] + 1 : 0 10 res = max(res, dp[i][j]) 11 } 12 } 13 return res 14 } 15 }
标签:count,LeetCode718,res,Maximum,Int,Repeated,var,repeating,dp 来源: https://www.cnblogs.com/strengthen/p/10509887.html