标签：nbsp img traints NPU val 技术 lis input into
Given a 2D grid of size n * m and an integer k. You need to shift the grid k times.
In one shift operation:
grid[i][j] becomes at grid[i][j + 1].grid[i][m - 1] becomes at grid[i + 1][0].grid[n - 1][m - 1] becomes at grid[0][0].Return the 2D grid after applying shift operation k times.
Example 1:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 1
Output: [[9,1,2],[3,4,5],[6,7,8]]
Example 2:
Input: grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]], k = 4
Output: [[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]
Example 3:
Input: grid = [[1,2,3],[4,5,6],[7,8,9]], k = 9
Output: [[1,2,3],[4,5,6],[7,8,9]]
Constraints:
1 <= grid.length <= 501 <= grid[i].length <= 50-1000 <= grid[i][j] <= 10000 <= k <= 100数学题，cnm
首先是最小化k，从原理（画个2*2）可以得到k可以优化到k % (n*m).
接下来要确定每个元素要转移到的位置，首先是column因为比较简单，算当前col（j）+ k = p，然后p < m时可以直接转换temp[i][p] = grid[i][j];
如果p>m, column就要再模m，p % m = j。
然后是比较难的row，如果p < m，就可以直接用temp[i][p] = grid[i][j]
如果p>m, 根据题意我们需要把每个第一列的元素下移，最后一个元素登顶,因为最终决定位置的是列，所以最重要加的只有(j+k) / m，最后再+row（i）模n
temp[(i + r) % n][p % m] = grid[i][j];
class Solution { public static List<List<Integer>> shiftGrid(int[][] grid, int k) { int[][] temp = new int[grid.length][grid[0].length]; // took temp grid int n = grid.length; int m = grid[0].length; int mod = n * m; k = k % mod; // minimize the k value for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { int p = j + k; // defines which col int r = p / (m); // defines which row，因为整个process是column在移动所以移动整数倍m是一样的 if (p < m) { temp[i][p] = grid[i][j]; } else { temp[(i + r) % n][p % m] = grid[i][j]; } } } // making temp grid into list List<List<Integer>> result = new LinkedList<>(); for (int[] t : temp) { LinkedList<Integer> c = new LinkedList<>(); for (int i : t) { c.add(i); } result.add(c); } return result; } }
class Solution { public static List<List<Integer>> shiftGrid(int[][] grid, int k) { int[][] temp = new int[grid.length][grid[0].length]; // took temp grid int n = grid.length; int m = grid[0].length; int mod = n * m; k = k % mod; // minimize the k value for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { int p = j + k; // defines which col int r = p / m; // defines which row temp[(i + r) % n][p % m] = grid[i][j]; } } // making temp grid into list List<List<Integer>> result = new LinkedList<>(); for (int[] t : temp) { LinkedList<Integer> c = new LinkedList<>(); for (int i : t) { c.add(i); } result.add(c); } return result; } }
标签：nbsp img traints NPU val 技术 lis input into
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11886518.html
