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

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. 1 <= A.length <= 30000
2. 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. 1 <= A.length <= 30000
2. 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 : true
11             if sum == (A[i] == 1)
12             {
13                 if i + K > n {return -1}
14                 f[i + K] = !f[i + K]
15                 sum = !sum
16                 ans += 1
17             }
18         }
19         return ans        
20     }
21 }