Leetcode995 Minimum Number of K Consecutive Bit Flips

##题目描述
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 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:

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

Example 2:

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Input: A = [1,1,0], K = 2
Output: -1
Explanation: No matter how we flip subarrays of size 2, we can't make the array become [1,1,1].

Example 3:

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

##思路
思路：先找到第一个0，找到第一个0以后，如果第一个0的位置i，i+k>当前数组的长度，则返回-1. 反之，在K循环里，对当前值进行按位异或(^=)且轮数加一。最后返回轮数。
##代码

class Solution {
    public int minKBitFlips(int[] A, int K) {
        int result=0;
        int i=0;
        while(i<A.length){
            if(A[i]==0){
                if(i+K>A.length)
                    return -1;
                else{
                    for(int j=0;j<K;j++){
                        A[i+j]^=1;
                    }
                    result++;
                }
            }else
                i++;
        }
        return result;
    }
}