LeetCode 1007. Minimal domino rotation with equal rows

1. Title

In a row of dominoes, A[i] and B[i] represent the upper and lower half of the i-th domino, respectively. (A domino is formed by tiling two numbers from 1 to 6 in the same column-there is a number on each half of the tile.)

We can rotate the i-th domino so that the values ​​of A[i] and B[i] are swapped.

Can return all the values of A or B, all the values are the same minimum number of revolutions .

If it cannot be done, return -1.

Example 1:

输入：A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
输出：2
解释：
图一表示：在我们旋转之前， A 和 B 给出的多米诺牌。
如果我们旋转第二个和第四个多米诺骨牌，
我们可以使上面一行中的每个值都等于 2，如图二所示。

示例 2：
输入：A = [3,5,1,2,3], B = [3,6,3,3,4]
输出：-1
解释：
在这种情况下，不可能旋转多米诺牌使一行的值相等。
 
提示：
1 <= A[i], B[i] <= 6
2 <= A.length == B.length <= 20000

Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/minimum-domino-rotations-for-equal-row The
copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.

2. Problem solving

• Find the number x where the number >= n
• Check the two numbers in each position:

1. All equal to x, no need to change, record times r2
2. Not equal to x, not satisfying the meaning of the question
3. One is equal to x, and the
   answer to record the number of rotations r1 is $\min (r1,n-r2-r1)$

class Solution {
    
    
public:
    int minDominoRotations(vector<int>& A, vector<int>& B) {
    
    
        int n = A.size();
        vector<int> count(7, 0);
        for(int i = 0; i < n; i++) 
        {
    
    
            count[A[i]]++;//计数
            count[B[i]]++;
        }
        int num = -1;
        for(int i = 1; i <= 6; ++i)
        {
    
    
            if(count[i] >= n)//个数达标的数
                num = i;
        }
        if(num == -1)
            return -1;
        int rotation = 0, notrotation = 0;
        for(int i = 0; i < n; ++i)
        {
    
    
            if(A[i] == num && B[i] == num)
                notrotation++;//两个都是，不需要交换
            else if(A[i] != num && B[i] != num)
                return -1;//都不等，不存在
            else if(A[i] == num)
                rotation++;
        }
        return min(rotation, n-notrotation-rotation);
    }
};

304 ms 100.7 MB

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