[Swift Weekly Contest 124]LeetCode995. K 连续位的最小翻转次数 | Minimum Number of K Consecutive Bit Flips

[Swift Weekly Contest 124]LeetCode995. K 连续位的最小翻转次数 | Minimum Number of K Consecutive Bit Flips,第1张

概述In an array A containing only 0s and 1s, a K-bit flip consists of choosing a (contiguous) subarray of length K and simultaneously changing every 0 in the subarray to 1, and every 1 in the subarray to

In an array A containing only 0s and 1s,a K-bit flip consists of choosing a (contiguous) subarray of length K and simultaneously changing every 0 in the subarray to 1,and every 1 in the subarray to 0.

Return the minimum number of K-bit flips required so that there is no 0 in the array.  If it is not possible,return -1.

Example 1:

input: A = [0,1,0],K = 1 Output: 2 Explanation: Flip A[0],then flip A[2]. 

Example 2:

input: A = [1,K = 2 Output: -1 Explanation: No matter how we flip subarrays of size 2,we can‘t make the array become [1,1]. 

Example 3:

input: A = [0,K = 3 Output: 3 Explanation: Flip A[0],A[1],A[2]: A becomes [1,0] Flip A[4],A[5],A[6]: A becomes [1,0] Flip A[5],A[6],A[7]: A becomes [1,1] 

Note:

1 <= A.length <= 30000 1 <= K <= A.length

在仅包含 0 和 1 的数组 A 中,一次 K 位翻转包括选择一个长度为 K的(连续)子数组,同时将子数组中的每个 0 更改为 1,而每个 1 更改为 0

返回所需的 K 位翻转的次数,以便数组没有值为 0 的元素。如果不可能,返回 -1。 

示例 1:

输入:A = [0,K = 1输出:2解释:先翻转 A[0],然后翻转 A[2]。

示例 2:

输入:A = [1,K = 2输出:-1解释:无论我们怎样翻转大小为 2 的子数组,我们都不能使数组变为 [1,1]。

示例 3:

输入:A = [0,K = 3输出:3解释:翻转 A[0],A[2]: A变成 [1,0]翻转 A[4],A[6]: A变成 [1,0]翻转 A[5],A[7]: A变成 [1,1] 

提示:

1 <= A.length <= 30000 1 <= K <= A.length Runtime: 752 ms Memory Usage: 19.1 MB
 1 class Solution { 2     func minKBitFlips(_ A: [Int],_ K: Int) -> Int { 3         var ans:Int = 0 4         var n:Int = A.count 5         var f:[Bool] = [Bool](repeating:false,count:n + 1) 6         var sum:Bool = false 7         for i in 0..<n 8         { 9             //异或10             sum = (sum == f[i]) ? false : true11             if sum == (A[i] == 1)12             {13                 if i + K > n {return -1}14                 f[i + K] = !f[i + K]15                 sum = !sum16                 ans += 117             }18         }19         return ans        20     }21 }
总结

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