1007. Minimum Domino Rotations For Equal Row

https://leetcode.com/problems/minimum-domino-rotations-for-equal-row/

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

We may rotate thei-th domino, so thatA[i]andB[i]swap values.

Return the minimum number of rotations so that all the values inAare the same, or all the values inB 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.

Solution

两排六点骰子，问最少swap次数能使得其中一排数字相同。

1. 方法一: 依次遍历1~6，看是否能满足

2. 法二：如果能使得两行数字相同，那数字必须是A[0]或B[0]。所以依次对A[i]和B[i]查是否等于A[0]、B[0]。如果A[0]满足，就不需要查B[0]了，因为只有当A[0] == B[0]时才可能A和B都满足条件。‌数个数时不能数==A[0]或B[0]的，因为A和B可能存在相同元素。

class Solution {
public:
    int minDominoRotations(vector<int>& A, vector<int>& B) {
        int res = INT_MAX;
        for (int i = 1; i <= 6; ++i) {
            int ac = 0, bc = 0;
            bool f = true;
            for (int j = 0; j < A.size(); ++j) {
                if (A[j] == i) ++ac;
                if (B[j] == i) ++bc;
                if (!(A[j] == i || B[j] == i)) {
                    f = false;
                    break;
                }
            }
            if (f) {
                res = min(res, ((int)A.size()) - max(ac, bc));
            }
        }
        if (res == INT_MAX) return -1;
        return res;
    }
};

class Solution {
public:
    int minDominoRotations(vector<int>& A, vector<int>& B) {
        if (A.size() != B.size()) return -1;
        auto n = A.size();
        for (int i = 0, a = 0, b = 0; i < n && (A[i] == A[0] || B[i] == A[0]); ++i) {
            if (A[i] != A[0]) ++a;
            if (B[i] != A[0]) ++b;
            if (i == A.size() - 1) return min(a, b); 
        } 
        for (int i = 0, a = 0, b = 0; i < n && (A[i] == B[0] || B[i] == B[0]); ++i) {
            if (A[i] != B[0]) ++a;
            if (B[i] != B[0]) ++b;
            if (i == n - 1) return min(a, b); 
        } 

        return -1;
    }
};