标签：dom imu new sam ann from equal for cannot
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.
class Solution { public int minDominoRotations(int[] A, int[] B) { int[] countA = new int[7]; int[] countB = new int[7]; int[] same = new int[7]; int n = A.length; for (int i = 0; i < n; i++) { countA[A[i]] += 1; countB[B[i]] += 1; if (A[i] == B[i]) { same[A[i]] += 1; } } for (int i = 1; i <= 6; i++) { if (countA[i] + countB[i] - same[i] == n) { return n - Math.max(countA[i], countB[i]); } } return -1; } }
[LC] 1007. Minimum Domino Rotations For Equal Row
标签：dom imu new sam ann from equal for cannot
原文地址：https://www.cnblogs.com/xuanlu/p/12294040.html
