980. Unique Paths III
선택한 UI 언어에 맞게 문제 텍스트를 러시아어에서 번역합니다. 코드는 변경하지 않습니다.
Вам дан 정수 배열 grid размером m x n, где grid[i][j] может быть:
1, представляющая начальную клетку. Существует ровно одна начальная клетка.
2, представляющая конечную клетку. Существует ровно одна конечная клетка.
0, представляющая пустые клетки, по которым можно ходить.
-1, представляющая препятствия, по которым нельзя ходить.
return количество 4-направленных путей от начальной клетки до конечной клетки, которые проходят по каждой непересекаемой клетке ровно один раз.
예제:
Input: grid = [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths:
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)
C# 해법
매칭됨/원본public class Solution {
private int rows, cols;
private int[,] grid;
private int pathCount;
private void Backtrack(int row, int col, int remain) {
if (grid[row, col] == 2 && remain == 1) {
pathCount += 1;
return;
}
int temp = grid[row, col];
grid[row, col] = -4;
remain -= 1;
int[] rowOffsets = {0, 0, 1, -1};
int[] colOffsets = {1, -1, 0, 0};
for (int i = 0; i < 4; ++i) {
int nextRow = row + rowOffsets[i];
int nextCol = col + colOffsets[i];
if (0 > nextRow || nextRow >= rows || 0 > nextCol || nextCol >= cols)
continue;
if (grid[nextRow, nextCol] < 0)
continue;
Backtrack(nextRow, nextCol, remain);
}
grid[row, col] = temp;
}
public int UniquePathsIII(int[,] grid) {
int nonObstacles = 0, startRow = 0, startCol = 0;
rows = grid.GetLength(0);
cols = grid.GetLength(1);
this.grid = grid;
for (int row = 0; row < rows; ++row)
for (int col = 0; col < cols; ++col) {
int cell = grid[row, col];
if (cell >= 0)
nonObstacles += 1;
if (cell == 1) {
startRow = row;
startCol = col;
}
}
pathCount = 0;
Backtrack(startRow, startCol, nonObstacles);
return pathCount;
}
}C++ 해법
자동 초안, 제출 전 검토#include <bits/stdc++.h>
using namespace std;
// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
private int rows, cols;
private int[,] grid;
private int pathCount;
private void Backtrack(int row, int col, int remain) {
if (grid[row, col] == 2 && remain == 1) {
pathCount += 1;
return;
}
int temp = grid[row, col];
grid[row, col] = -4;
remain -= 1;
vector<int>& rowOffsets = {0, 0, 1, -1};
vector<int>& colOffsets = {1, -1, 0, 0};
for (int i = 0; i < 4; ++i) {
int nextRow = row + rowOffsets[i];
int nextCol = col + colOffsets[i];
if (0 > nextRow || nextRow >= rows || 0 > nextCol || nextCol >= cols)
continue;
if (grid[nextRow, nextCol] < 0)
continue;
Backtrack(nextRow, nextCol, remain);
}
grid[row, col] = temp;
}
public int UniquePathsIII(int[,] grid) {
int nonObstacles = 0, startRow = 0, startCol = 0;
rows = grid.GetLength(0);
cols = grid.GetLength(1);
this.grid = grid;
for (int row = 0; row < rows; ++row)
for (int col = 0; col < cols; ++col) {
int cell = grid[row, col];
if (cell >= 0)
nonObstacles += 1;
if (cell == 1) {
startRow = row;
startCol = col;
}
}
pathCount = 0;
Backtrack(startRow, startCol, nonObstacles);
return pathCount;
}
}Java 해법
매칭됨/원본class Solution {
int rows, cols;
int[][] grid;
int path_count;
protected void backtrack(int row, int col, int remain) {
if (this.grid[row][col] == 2 && remain == 1) {
this.path_count += 1;
return;
}
int temp = grid[row][col];
grid[row][col] = -4;
remain -= 1;
int[] row_offsets = {0, 0, 1, -1};
int[] col_offsets = {1, -1, 0, 0};
for (int i = 0; i < 4; ++i) {
int next_row = row + row_offsets[i];
int next_col = col + col_offsets[i];
if (0 > next_row || next_row >= this.rows || 0 > next_col || next_col >= this.cols)
continue;
if (grid[next_row][next_col] < 0)
continue;
backtrack(next_row, next_col, remain);
}
grid[row][col] = temp;
}
public int uniquePathsIII(int[][] grid) {
int non_obstacles = 0, start_row = 0, start_col = 0;
this.rows = grid.length;
this.cols = grid[0].length;
for (int row = 0; row < rows; ++row)
for (int col = 0; col < cols; ++col) {
int cell = grid[row][col];
if (cell >= 0)
non_obstacles += 1;
if (cell == 1) {
start_row = row;
start_col = col;
}
}
this.path_count = 0;
this.grid = grid;
backtrack(start_row, start_col, non_obstacles);
return this.path_count;
}
}Python 해법
매칭됨/원본class Solution:
def uniquePathsIII(self, grid: list[list[int]]) -> int:
def backtrack(row, col, remain):
if grid[row][col] == 2 and remain == 1:
self.path_count += 1
return
temp = grid[row][col]
grid[row][col] = -4
remain -= 1
for ro, co in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
next_row, next_col = row + ro, col + co
if 0 <= next_row < self.rows and 0 <= next_col < self.cols and grid[next_row][next_col] >= 0:
backtrack(next_row, next_col, remain)
grid[row][col] = temp
non_obstacles = 0
start_row = start_col = 0
self.rows, self.cols = len(grid), len(grid[0])
for row in range(self.rows):
for col in range(self.cols):
if grid[row][col] >= 0:
non_obstacles += 1
if grid[row][col] == 1:
start_row, start_col = row, col
self.path_count = 0
backtrack(start_row, start_col, non_obstacles)
return self.path_countGo 해법
매칭됨/원본type Solution struct {
rows, cols int
grid [][]int
pathCount int
}
func (s *Solution) backtrack(row, col, remain int) {
if s.grid[row][col] == 2 && remain == 1 {
s.pathCount++
return
}
temp := s.grid[row][col]
s.grid[row][col] = -4
remain--
rowOffsets := []int{0, 0, 1, -1}
colOffsets := []int{1, -1, 0, 0}
for i := 0; i < 4; i++ {
nextRow := row + rowOffsets[i]
nextCol := col + colOffsets[i]
if nextRow < 0 || nextRow >= s.rows || nextCol < 0 || nextCol >= s.cols {
continue
}
if s.grid[nextRow][nextCol] < 0 {
continue
}
s.backtrack(nextRow, nextCol, remain)
}
s.grid[row][col] = temp
}
func (s *Solution) UniquePathsIII(grid [][]int) int {
nonObstacles, startRow, startCol := 0, 0, 0
s.rows = len(grid)
s.cols = len(grid[0])
for row := 0; row < s.rows; row++ {
for col := 0; col < s.cols; col++ {
cell := grid[row][col]
if cell >= 0 {
nonObstacles++
}
if cell == 1 {
startRow = row
startCol = col
}
}
}
s.pathCount = 0
s.grid = grid
s.backtrack(startRow, startCol, nonObstacles)
return s.pathCount
}Algorithm
1⃣Как видно, метод обратного отслеживания (backtracking) является методологией для решения определенного типа задач.
2⃣Для задачи обратного отслеживания можно сказать, что существует тысяча реализаций обратного отслеживания на тысячу людей, как будет видно из дальнейшей реализации.
3⃣Здесь мы просто покажем один 예제 реализации, следуя псевдокоду, показанному в разделе интуиции.
😎
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