980. Unique Paths III
El texto de la tarea se traduce del ruso para el idioma seleccionado. El código no cambia.
Вам дан entero arreglo 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.
Ejemplo:
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# solución
coincidente/originalpublic 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++ solución
borrador automático, revisar antes de enviar#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 solución
coincidente/originalclass 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 solución
coincidente/originalclass 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 solución
coincidente/originaltype 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 Ejemplo реализации, следуя псевдокоду, показанному в разделе интуиции.
😎
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