正在初始化 std::memset
Initializing std::memset
我正在使用 std::memset
解决此处复制的 LeetCode problem。目前,map_cherries
已被初始化为:
map_cherries[70][70][70] = {}
...
std::memset(map_cherries, -1, sizeof(map_cherries));
有没有办法同时设置 map_cherries
的维度和值,或者可能有任何 better/alternative 方法来做到这一点?谢谢!
问题
Given a rows x cols
matrix grid representing a field of cherries.
Each cell in grid represents the number of cherries that you can
collect.
You have two robots that can collect cherries for you, Robot #1 is
located at the top-left corner (0,0)
, and Robot #2 is located at the
top-right corner (0, cols-1)
of the grid.
Return the maximum number of cherries collection using both robots by
following the rules below:
- From a cell
(i,j)
, robots can move to cell (i+1, j-1)
, (i+1, j)
or (i+1, j+1)
.
- When any robot is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).
- When both robots stay on the same cell, only one of them takes the cherries.
- Both robots cannot move outside of the grid at any moment.
- Both robots should reach the bottom row in the grid.
Example 1:
Input: grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24
Explanation: Path of robot #1 and #2 are described in color green and
blue respectively.
Cherries taken by Robot #1, (3 + 2 + 5 + 2) = 12
.
Cherries taken by Robot #2, (1 + 5 + 5 + 1) = 12
.
Total of cherries: 12 + 12 = 24
.
Example 2:
Input: grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
Output: 28
Explanation: Path of robot #1 and #2 are described in
color green and blue respectively.
Cherries taken by Robot #1, (1 + 9 + 5 + 2) = 17
.
Cherries taken by Robot #2, (1 + 3 + 4 + 3) = 11
.
Total of cherries: 17 + 11 = 28
.
Constraints:
rows == grid.length
cols == grid[i].length
2 <= rows, cols <= 70
0 <= grid[i][j] <= 100
尝试
#include <cstdint>
#include <vector>
#include <cstring>
struct Solution {
int map_cherries[70][70][70] = {};
int cherryPickup(std::vector<std::vector<int>>& grid) {
std::memset(map_cherries, -1, sizeof(map_cherries));
const std::size_t row_length = grid.size();
const std::size_t col_length = grid[0].size();
return depth_first_search(grid, row_length, col_length, 0, 0, col_length - 1);
}
private:
int depth_first_search(
std::vector<std::vector<int>>& grid,
const std::size_t row_length,
const std::size_t col_length,
int row,
int left_robot,
int right_robot
) {
if (row == row_length) {
return 0;
}
if (map_cherries[row][left_robot][right_robot] != -1) {
return map_cherries[row][left_robot][right_robot];
}
int max_cherries = 0;
for (int left = -1; left < 2; left++) {
for (int right = -1; right < 2; right++) {
const int curr_left_robot = left_robot + left;
const int curr_right_robot = right_robot + right;
if (curr_left_robot > -1 and curr_left_robot < col_length and curr_right_robot > -1 and curr_right_robot < col_length) {
max_cherries = std::max(max_cherries, depth_first_search(
grid,
row_length,
col_length,
row + 1,
curr_left_robot,
curr_right_robot
));
}
}
}
int total_cherries = grid[row][left_robot];
if (left_robot != right_robot) {
total_cherries += grid[row][right_robot];
}
total_cherries += max_cherries;
return map_cherries[row][left_robot][right_robot] = total_cherries;
}
};
输入
[[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
[[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
[[1,10,0,3,86,40],[0,0,0,3,86,40],[0,0,3,3,86,40],[9,0,3,3,86,40], [1,0,40,3,86,40],[0,22,0,3,86,40],[99,0,3,3,86,40],[9,0,3,3,86,40]]
产出
24
22
819
参考资料
一种方法是这样的:
auto map_cherries = std::vector(70, std::vector(70, std::vector(70, -1)));
只能从 c++17 开始使用。
我正在使用 std::memset
解决此处复制的 LeetCode problem。目前,map_cherries
已被初始化为:
map_cherries[70][70][70] = {}
...
std::memset(map_cherries, -1, sizeof(map_cherries));
有没有办法同时设置 map_cherries
的维度和值,或者可能有任何 better/alternative 方法来做到这一点?谢谢!
问题
Given a
rows x cols
matrix grid representing a field of cherries. Each cell in grid represents the number of cherries that you can collect.You have two robots that can collect cherries for you, Robot #1 is located at the top-left corner
(0,0)
, and Robot #2 is located at the top-right corner(0, cols-1)
of the grid.Return the maximum number of cherries collection using both robots by following the rules below:
- From a cell
(i,j)
, robots can move to cell(i+1, j-1)
,(i+1, j)
or(i+1, j+1)
.- When any robot is passing through a cell, It picks it up all cherries, and the cell becomes an empty cell (0).
- When both robots stay on the same cell, only one of them takes the cherries.
- Both robots cannot move outside of the grid at any moment.
- Both robots should reach the bottom row in the grid.
Example 1:
Input:
grid = [[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
Output: 24Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1,(3 + 2 + 5 + 2) = 12
.Cherries taken by Robot #2,
(1 + 5 + 5 + 1) = 12
. Total of cherries:12 + 12 = 24
.Example 2:
Input:
grid = [[1,0,0,0,0,0,1],[2,0,0,0,0,3,0],[2,0,9,0,0,0,0],[0,3,0,5,4,0,0],[1,0,2,3,0,0,6]]
Output: 28
Explanation: Path of robot #1 and #2 are described in color green and blue respectively.
Cherries taken by Robot #1,(1 + 9 + 5 + 2) = 17
.Cherries taken by Robot #2,
(1 + 3 + 4 + 3) = 11
.Total of cherries:
17 + 11 = 28
.Constraints:
rows == grid.length
cols == grid[i].length
2 <= rows, cols <= 70
0 <= grid[i][j] <= 100
尝试
#include <cstdint>
#include <vector>
#include <cstring>
struct Solution {
int map_cherries[70][70][70] = {};
int cherryPickup(std::vector<std::vector<int>>& grid) {
std::memset(map_cherries, -1, sizeof(map_cherries));
const std::size_t row_length = grid.size();
const std::size_t col_length = grid[0].size();
return depth_first_search(grid, row_length, col_length, 0, 0, col_length - 1);
}
private:
int depth_first_search(
std::vector<std::vector<int>>& grid,
const std::size_t row_length,
const std::size_t col_length,
int row,
int left_robot,
int right_robot
) {
if (row == row_length) {
return 0;
}
if (map_cherries[row][left_robot][right_robot] != -1) {
return map_cherries[row][left_robot][right_robot];
}
int max_cherries = 0;
for (int left = -1; left < 2; left++) {
for (int right = -1; right < 2; right++) {
const int curr_left_robot = left_robot + left;
const int curr_right_robot = right_robot + right;
if (curr_left_robot > -1 and curr_left_robot < col_length and curr_right_robot > -1 and curr_right_robot < col_length) {
max_cherries = std::max(max_cherries, depth_first_search(
grid,
row_length,
col_length,
row + 1,
curr_left_robot,
curr_right_robot
));
}
}
}
int total_cherries = grid[row][left_robot];
if (left_robot != right_robot) {
total_cherries += grid[row][right_robot];
}
total_cherries += max_cherries;
return map_cherries[row][left_robot][right_robot] = total_cherries;
}
};
输入
[[3,1,1],[2,5,1],[1,5,5],[2,1,1]]
[[1,0,0,3],[0,0,0,3],[0,0,3,3],[9,0,3,3]]
[[1,10,0,3,86,40],[0,0,0,3,86,40],[0,0,3,3,86,40],[9,0,3,3,86,40], [1,0,40,3,86,40],[0,22,0,3,86,40],[99,0,3,3,86,40],[9,0,3,3,86,40]]
产出
24
22
819
参考资料
一种方法是这样的:
auto map_cherries = std::vector(70, std::vector(70, std::vector(70, -1)));
只能从 c++17 开始使用。