📅  最后修改于: 2023-12-03 15:41:01.786000             🧑  作者: Mango
在给定的矩阵中,我们需要计算出所有值等于给定值 x 的单元格的数量,然后将其按照行和列的方式进行展示。这个问题可以使用暴力方法和优化方法来解决。
暴力方法有两个步骤:
这个方法遍历了整个矩阵,时间复杂度为 O(n^2),其中 n 是矩阵的长度和宽度。
优化方法仅需要遍历一次矩阵,并且时间复杂度为 O(n)。
def count_cell(matrix, x):
cell_count = 0
row_count = [0] * len(matrix)
col_count = [0] * len(matrix[0])
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == x:
cell_count += 1
row_count[i] += 1
col_count[j] += 1
result = []
for i in range(len(matrix)):
row_sum = 0
for j in range(len(matrix[0])):
row_sum += matrix[i][j]
result.append(row_sum + row_count[i])
for j in range(len(matrix[0])):
col_sum = 0
for i in range(len(matrix)):
col_sum += matrix[i][j]
result.append(col_sum + col_count[j])
return result
def count_cell_optimized(matrix, x):
cell_count = 0
row_count = [0] * len(matrix)
col_count = [0] * len(matrix[0])
for i in range(len(matrix)):
for j in range(len(matrix[0])):
if matrix[i][j] == x:
cell_count += 1
row_count[i] += 1
col_count[j] += 1
result = []
for i in range(len(matrix)):
row_sum = sum(matrix[i])
result.append(row_sum + row_count[i])
for j in range(len(matrix[0])):
col_sum = sum(matrix[i][j] for i in range(len(matrix)))
result.append(col_sum + col_count[j])
return result
function countCell(matrix, x) {
let cellCount = 0;
let rowCount = Array(matrix.length).fill(0);
let colCount = Array(matrix[0].length).fill(0);
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] === x) {
cellCount++;
rowCount[i]++;
colCount[j]++;
}
}
}
let result = [];
for (let i = 0; i < matrix.length; i++) {
let rowSum = 0;
for (let j = 0; j < matrix[0].length; j++) {
rowSum += matrix[i][j];
}
result.push(rowSum + rowCount[i]);
}
for (let j = 0; j < matrix[0].length; j++) {
let colSum = 0;
for (let i = 0; i < matrix.length; i++) {
colSum += matrix[i][j];
}
result.push(colSum + colCount[j]);
}
return result;
}
function countCellOptimized(matrix, x) {
let cellCount = 0;
let rowCount = Array(matrix.length).fill(0);
let colCount = Array(matrix[0].length).fill(0);
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] === x) {
cellCount++;
rowCount[i]++;
colCount[j]++;
}
}
}
let result = [];
for (let i = 0; i < matrix.length; i++) {
let rowSum = matrix[i].reduce((total, currentValue) => total + currentValue);
result.push(rowSum + rowCount[i]);
}
for (let j = 0; j < matrix[0].length; j++) {
let colSum = 0;
for (let i = 0; i < matrix.length; i++) {
colSum += matrix[i][j];
}
result.push(colSum + colCount[j]);
}
return result;
}
矩阵中等于 X 的单元格计数构造为行和列的总和问题可以使用暴力方法和优化方法来解决。其中,优化方法遍历一次矩阵,并将时间复杂度从 O(n^2) 优化到了 O(n)。