獨立站都有哪些百度快速排名提升
題目:
給你一個下標從 0 開始、大小為 n * m 的二維整數(shù)矩陣 grid ,定義一個下標從 0 開始、大小為 n * m 的的二維矩陣 p。如果滿足以下條件,則稱 p 為 grid 的 乘積矩陣 :
對于每個元素 p[i][j] ,它的值等于除了 grid[i][j] 外所有元素的乘積。乘積對 12345 取余數(shù)。
返回 grid 的乘積矩陣。
示例 1:
輸入:grid = [[1,2],[3,4]]
輸出:[[24,12],[8,6]]
解釋:p[0][0] = grid[0][1] * grid[1][0] * grid[1][1] = 2 * 3 * 4 = 24
p[0][1] = grid[0][0] * grid[1][0] * grid[1][1] = 1 * 3 * 4 = 12
p[1][0] = grid[0][0] * grid[0][1] * grid[1][1] = 1 * 2 * 4 = 8
p[1][1] = grid[0][0] * grid[0][1] * grid[1][0] = 1 * 2 * 3 = 6
所以答案是 [[24,12],[8,6]] 。
示例 2:
輸入:grid = [[12345],[2],[1]]
輸出:[[2],[0],[0]]
解釋:p[0][0] = grid[0][1] * grid[0][2] = 2 * 1 = 2
p[0][1] = grid[0][0] * grid[0][2] = 12345 * 1 = 12345. 12345 % 12345 = 0 ,所以 p[0][1] = 0
p[0][2] = grid[0][0] * grid[0][1] = 12345 * 2 = 24690. 24690 % 12345 = 0 ,所以 p[0][2] = 0
所以答案是 [[2],[0],[0]] 。
提示:
1 <= n == grid.length <= 10^5
1 <= m == grid[i].length <= 10^5
2 <= n * m <= 10^5
1 <= grid[i][j] <= 10^9
java代碼:
class Solution {public int[][] constructProductMatrix(int[][] grid) {final int MOD = 12345;int n = grid.length, m = grid[0].length;int[][] p = new int[n][m];long suf = 1; // 后綴乘積for (int i = n - 1; i >= 0; i--) {for (int j = m - 1; j >= 0; j--) {p[i][j] = (int) suf; // p[i][j] 先初始化成后綴乘積suf = suf * grid[i][j] % MOD;}}long pre = 1; // 前綴乘積for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {p[i][j] = (int) (p[i][j] * pre % MOD); // 然后再乘上前綴乘積pre = pre * grid[i][j] % MOD;}}return p;}
}