Given an array A
of 0s and 1s,we may change up to K
values from 0 to 1.
Return the length of the longest (contiguous) subarray that contains only 1s.
Example 1:
Input: A = [1,1,0],K = 2 Output: 6 Explanation: [1,1,1] Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Example 2:
Input: A = [0,1],K = 3 Output: 10 Explanation: [0,1,1] Bolded numbers were flipped from 0 to 1. The longest subarray is underlined.
Note:
1 <= A.length <= 20000
0 <= K <= A.length
-
A[i]
is0
or1
给定一个由若干 0
和 1
组成的数组 A
,我们最多可以将 K
个值从 0 变成 1 。
返回仅包含 1 的最长(连续)子数组的长度。
示例 1:
输入:A = [1,K = 2 输出:6 解释: [1,1] 粗体数字从 0 翻转到 1,最长的子数组长度为 6。
示例 2:
输入:A = [0,K = 3 输出:10 解释: [0,1] 粗体数字从 0 翻转到 1,最长的子数组长度为 10。
提示:
1 <= A.length <= 20000
0 <= K <= A.length
-
A[i]
为0
或1
Runtime: 476 ms
Memory Usage: 18.9 MB
1 class Solution { 2 func longestOnes(_ A: [Int],_ K: Int) -> Int { 3 var res:Int = 0 4 var zero:Int = 0 5 var left:Int = 0 6 for right in 0..<A.count 7 { 8 if A[right] == 0 9 { 10 zero += 1 11 } 12 while (zero > K) 13 { 14 if A[left] == 0 15 { 16 zero -= 1 17 } 18 left += 1 19 } 20 res = max(res,right - left + 1) 21 } 22 return res 23 } 24 }
Runtime: 576 ms
Memory Usage: 18.9 MB
1 class Solution { 2 func longestOnes(_ A: [Int],_ K: Int) -> Int { 3 var n:Int = A.count 4 var pre:[Int] = [Int](repeating:0,count:n) 5 for i in 0..<n 6 { 7 if A[i] == 0 8 { 9 pre[i] = 1 10 } 11 } 12 for i in 1..<n 13 { 14 pre[i] = pre[i - 1] + pre[i] 15 } 16 var fans:Int = 0 17 for i in -1..<(n - 1) 18 { 19 var lo:Int = i + 1 20 var hi:Int = n - 1 21 var ans:Int = i 22 while(lo <= hi) 23 { 24 var mid:Int = (lo + hi) / 2 25 var val:Int = pre[mid] 26 if i >= 0 27 { 28 val -= pre[i] 29 } 30 if val <= K 31 { 32 ans = mid 33 lo = mid + 1 34 } 35 else 36 { 37 hi = mid - 1 38 } 39 } 40 fans = max(fans,ans - i) 41 } 42 return fans 43 } 44 }
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