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102 lines
2.8 KiB
Markdown
Executable File
102 lines
2.8 KiB
Markdown
Executable File
# [1252. Cells with Odd Values in a Matrix](https://leetcode.com/problems/cells-with-odd-values-in-a-matrix/)
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## 题目
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Given `n` and `m` which are the dimensions of a matrix initialized by zeros and given an array `indices` where `indices[i] = [ri, ci]`. For each pair of `[ri, ci]` you have to increment all cells in row `ri` and column `ci` by 1.
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Return *the number of cells with odd values* in the matrix after applying the increment to all `indices`.
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**Example 1**:
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Input: n = 2, m = 3, indices = [[0,1],[1,1]]
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Output: 6
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Explanation: Initial matrix = [[0,0,0],[0,0,0]].
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After applying first increment it becomes [[1,2,1],[0,1,0]].
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The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
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**Example 2**:
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Input: n = 2, m = 2, indices = [[1,1],[0,0]]
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Output: 0
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Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
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**Constraints**:
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- `1 <= n <= 50`
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- `1 <= m <= 50`
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- `1 <= indices.length <= 100`
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- `0 <= indices[i][0] < n`
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- `0 <= indices[i][1] < m`
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## 题目大意
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给你一个 n 行 m 列的矩阵，最开始的时候，每个单元格中的值都是 0。另有一个索引数组 indices，indices[i] = [ri, ci] 中的 ri 和 ci 分别表示指定的行和列（从 0 开始编号）。你需要将每对 [ri, ci] 指定的行和列上的所有单元格的值加 1。请你在执行完所有 indices 指定的增量操作后，返回矩阵中 「奇数值单元格」 的数目。
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提示：
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- 1 <= n <= 50
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- 1 <= m <= 50
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- 1 <= indices.length <= 100
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- 0 <= indices[i][0] < n
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- 0 <= indices[i][1] < m
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## 解题思路
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- 给出一个 n * m 的矩阵，和一个数组，数组里面包含一些行列坐标，并在指定坐标上 + 1，问最后 n * m 的矩阵中奇数的总数。
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- 暴力方法按照题意模拟即可。
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## 代码
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```go
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package leetcode
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// 解法一 暴力法
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func oddCells(n int, m int, indices [][]int) int {
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matrix, res := make([][]int, n), 0
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for i := range matrix {
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matrix[i] = make([]int, m)
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}
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for _, indice := range indices {
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for i := 0; i < m; i++ {
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matrix[indice[0]][i]++
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}
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for j := 0; j < n; j++ {
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matrix[j][indice[1]]++
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}
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}
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for _, m := range matrix {
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for _, v := range m {
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if v&1 == 1 {
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res++
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}
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}
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}
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return res
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}
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// 解法二 暴力法
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func oddCells1(n int, m int, indices [][]int) int {
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rows, cols, count := make([]int, n), make([]int, m), 0
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for _, pair := range indices {
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rows[pair[0]]++
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cols[pair[1]]++
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}
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for i := 0; i < n; i++ {
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for j := 0; j < m; j++ {
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if (rows[i]+cols[j])%2 == 1 {
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count++
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}
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}
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}
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return count
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}
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``` |