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44 lines
1.2 KiB
44 lines
1.2 KiB
public class Solution1260 {
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public List<List<Integer>> shiftGrid(int[][] grid, int k) {
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// 行数
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int m = grid.length;
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// 列数
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int n = grid[0].length;
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// 偏移值，取下模
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k = k % (m * n);
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// 反向取下数量，因为我打算直接从头填充新的矩阵
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/*
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* 比如
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* 1 2 3
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* 4 5 6
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* 7 8 9
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* 需要变成
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* 9 1 2
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* 3 4 5
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* 6 7 8
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* 就要从 9 开始填充
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*/
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int reverseK = m * n - k;
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List<List<Integer>> matrix = new ArrayList<>();
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// 这类就是两层循环
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for (int i = 0; i < m; i++) {
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List<Integer> line = new ArrayList<>();
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for (int j = 0; j < n; j++) {
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// 数量会随着循环迭代增长, 确认是第几个
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int currentNum = reverseK + i * n + (j + 1);
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// 这里处理下到达矩阵末尾后减掉 m * n
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if (currentNum > m * n) {
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currentNum -= m * n;
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}
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// 根据矩阵列数 n 算出在原来矩阵的位置
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int last = (currentNum - 1) % n;
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int passLine = (currentNum - 1) / n;
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line.add(grid[passLine][last]);
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}
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matrix.add(line);
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}
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return matrix;
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}
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}
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