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English Version

题目描述

给你一个 m * n 的网格,其中每个单元格不是 0(空)就是 1(障碍物)。每一步,您都可以在空白单元格中上、下、左、右移动。

如果您 最多 可以消除 k 个障碍物,请找出从左上角 (0, 0) 到右下角 (m-1, n-1) 的最短路径,并返回通过该路径所需的步数。如果找不到这样的路径,则返回 -1 。

 

示例 1:

输入: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1
输出:6
解释:
不消除任何障碍的最短路径是 10。
消除位置 (3,2) 处的障碍后,最短路径是 6 。该路径是 (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).

示例 2:

输入:grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1
输出:-1
解释:我们至少需要消除两个障碍才能找到这样的路径。

 

提示:

  • grid.length == m
  • grid[0].length == n
  • 1 <= m, n <= 40
  • 1 <= k <= m*n
  • grid[i][j] 是 0 或 1
  • grid[0][0] == grid[m-1][n-1] == 0

解法

BFS 最短路模型。

对于本题,如果 k >= m + n - 3,那么最短路径长度一定是 m + n - 2,直接返回,无需 BFS 计算。

Python3

class Solution:
    def shortestPath(self, grid: List[List[int]], k: int) -> int:
        m, n = len(grid), len(grid[0])
        if k >= m + n - 3:
            return m + n - 2
        q = deque([(0, 0, k)])
        vis = {(0, 0, k)}
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                i, j, k = q.popleft()
                for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n:
                        if x == m - 1 and y == n - 1:
                            return ans
                        if grid[x][y] == 0 and (x, y, k) not in vis:
                            q.append((x, y, k))
                            vis.add((x, y, k))
                        if grid[x][y] == 1 and k > 0 and (x, y, k - 1) not in vis:
                            q.append((x, y, k - 1))
                            vis.add((x, y, k - 1))
        return -1

Java

class Solution {
    public int shortestPath(int[][] grid, int k) {
        int m = grid.length;
        int n = grid[0].length;
        if (k >= m + n - 3) {
            return m + n - 2;
        }
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(new int[] {0, 0, k});
        boolean[][][] vis = new boolean[m][n][k + 1];
        vis[0][0][k] = true;
        int ans = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            ++ans;
            for (int i = q.size(); i > 0; --i) {
                int[] p = q.poll();
                k = p[2];
                for (int j = 0; j < 4; ++j) {
                    int x = p[0] + dirs[j];
                    int y = p[1] + dirs[j + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        if (x == m - 1 && y == n - 1) {
                            return ans;
                        }
                        if (grid[x][y] == 0 && !vis[x][y][k]) {
                            q.offer(new int[] {x, y, k});
                            vis[x][y][k] = true;
                        } else if (grid[x][y] == 1 && k > 0 && !vis[x][y][k - 1]) {
                            q.offer(new int[] {x, y, k - 1});
                            vis[x][y][k - 1] = true;
                        }
                    }
                }
            }
        }
        return -1;
    }
}

C++

class Solution {
public:
    int shortestPath(vector<vector<int>>& grid, int k) {
        int m = grid.size(), n = grid[0].size();
        if (k >= m + n - 3) return m + n - 2;
        queue<vector<int>> q;
        q.push({0, 0, k});
        vector<vector<vector<bool>>> vis(m, vector<vector<bool>>(n, vector<bool>(k + 1)));
        vis[0][0][k] = true;
        int ans = 0;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            ++ans;
            for (int i = q.size(); i > 0; --i) {
                auto p = q.front();
                k = p[2];
                q.pop();
                for (int j = 0; j < 4; ++j) {
                    int x = p[0] + dirs[j], y = p[1] + dirs[j + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n) {
                        if (x == m - 1 && y == n - 1) return ans;
                        if (grid[x][y] == 0 && !vis[x][y][k]) {
                            q.push({x, y, k});
                            vis[x][y][k] = true;
                        } else if (grid[x][y] == 1 && k > 0 && !vis[x][y][k - 1]) {
                            q.push({x, y, k - 1});
                            vis[x][y][k - 1] = true;
                        }
                    }
                }
            }
        }
        return -1;
    }
};

Go

func shortestPath(grid [][]int, k int) int {
	m, n := len(grid), len(grid[0])
	if k >= m+n-3 {
		return m + n - 2
	}
	q := [][]int{[]int{0, 0, k}}
	vis := make([][][]bool, m)
	for i := range vis {
		vis[i] = make([][]bool, n)
		for j := range vis[i] {
			vis[i][j] = make([]bool, k+1)
		}
	}
	vis[0][0][k] = true
	dirs := []int{-1, 0, 1, 0, -1}
	ans := 0
	for len(q) > 0 {
		ans++
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			k = p[2]
			for j := 0; j < 4; j++ {
				x, y := p[0]+dirs[j], p[1]+dirs[j+1]
				if x >= 0 && x < m && y >= 0 && y < n {
					if x == m-1 && y == n-1 {
						return ans
					}
					if grid[x][y] == 0 && !vis[x][y][k] {
						q = append(q, []int{x, y, k})
						vis[x][y][k] = true
					} else if grid[x][y] == 1 && k > 0 && !vis[x][y][k-1] {
						q = append(q, []int{x, y, k - 1})
						vis[x][y][k-1] = true
					}
				}
			}
		}
	}
	return -1
}

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