LeetCode #1007. Minimum Domino Rotations For Equal Row

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本文探讨了一个算法问题，即如何通过最少的旋转次数使一排多米诺骨牌的顶部或底部数值全部相同。通过分析给定的两组数值，我们设计了一种解决方案，首先判断是否有可能实现目标，然后计算出达到目标所需的最小旋转次数。

题目描述：

In a row of dominoes, A[i] and B[i] represent the top and bottom halves of the i-th domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)

We may rotate the i-th domino, so that A[i] and B[i] swap values.

Return the minimum number of rotations so that all the values in A are the same, or all the values in B are the same.

If it cannot be done, return -1.

Example 1:

Input: A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
Output: 2
Explanation: 
The first figure represents the dominoes as given by A and B: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.

Example 2:

Input: A = [3,5,1,2,3], B = [3,6,3,3,4]
Output: -1
Explanation: 
In this case, it is not possible to rotate the dominoes to make one row of values equal.

Note:

1 <= A[i], B[i] <= 6
2 <= A.length == B.length <= 20000

class Solution {
public:
    int minDominoRotations(vector<int>& A, vector<int>& B) {
        // 最终结果必然是让A或B全部变成A[0]或B[0]，那么令这两个数作为候选
        int a=A[0], b=B[0];
        for(int i=1;i<A.size();i++)
        {
            if(A[i]!=a&&B[i]!=a) a=-1;
            if(A[i]!=b&&B[i]!=b) b=-1;
            if(a==-1&&b==-1) return -1;
        }
        
        int c=a!=-1?a:b;
        int count1=0, count2=0;
        for(int i=0;i<A.size();i++)
        {
            if(A[i]!=c) count1++;
            if(B[i]!=c) count2++;
        }
        return min(count1,count2);
    }
};