980. Unique Paths III
Le texte du problème est traduit du russe pour la langue sélectionnée. Le code reste inchangé.
Вам дан entier tableau 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.
Exemple:
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# solution
correspondant/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++ solution
brouillon automatique, à relire avant soumission#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 solution
correspondant/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 solution
correspondant/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 solution
correspondant/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 Exemple реализации, следуя псевдокоду, показанному в разделе интуиции.
😎
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