[LeetCode] 1162. As Far from Land as Possible 地圖分析
Given ann x ngridcontaining only values0and1, where0represents waterand1represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance.If no land or water exists in the grid, return-1.
The distance used in this problem is the Manhattan distance:the distance between two cells(x0, y0)
(x1, y1)is|x0 - x1| + |y0 - y1|.
Example 1:
Input: grid = [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation: The cell (1, 1) is as far as possible from all the land with distance 2.
Example 2:
Input: grid = [[1,0,0],[0,0,0],[0,0,0]] Output: 4 Explanation: The cell (2, 2) is as far as possible from all the land with distance 4.
Constraints:
n == grid.lengthn == grid[i].length1 <= n<= 100grid[i][j]is0or1
這道題給了一個只有0和1的二維陣列，說是0表示水，1表示陸地，現在讓找出一個0的位置，使得其離最近的1的距離最大，這裡的距離用曼哈頓距離表示。這裡最暴力的方法就是遍歷每個0的位置，對於每個遍歷到的0，再遍歷每個1的位置，計算它們的距離，找到最小的距離儲存為該0位置的距離，然後在所有0位置的距離中找出最大的。這種方法不是很高效，目測無法通過 OJ，博主都沒有嘗試。其實這道題的比較好的解法是建立距離場，即每個大於1的數字表示該位置到任意一個1位置的最短距離，在之前那道
解法一：
class Solution {
public:
int maxDistance(vector<vector<int>>& grid) {
int step = 0, n = grid.size();
vector<vector<int>> dirs{{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
queue<vector<int>> q;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 0) continue;
q.push(vector<int>{i, j});
}
}
if (q.size() == 0 || q.size() == n * n) return -1;
while (!q.empty()) {
++step;
for (int k = q.size(); k > 0; --k) {
auto t = q.front(); q.pop();
for (auto dir : dirs) {
int x = t[0] + dir[0], y = t[1] + dir[1];
if (x < 0 || x >= n || y < 0 || y >= n || grid[x][y] != 0) continue;
grid[x][y] = step;
q.push({x, y});
}
}
}
return step - 1;
}
};
我們也可以強行用 DFS 來做，這裡對於每一個值為1的點都呼叫遞迴函式，更新跟其相連的所有0位置的距離，最終也能算出整個距離場，從而查找出最大的距離，參見程式碼如下：
解法二：
class Solution {
public:
int maxDistance(vector<vector<int>>& grid) {
int res = -1, n = grid.size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
grid[i][j] = 0;
helper(grid, i, j);
}
}
}
for (int i = n - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
if (grid[i][j] > 1) res = max(res, grid[i][j] - 1);
}
}
return res;
}
void helper(vector<vector<int>>& grid, int i, int j, int dist = 1) {
int n = grid.size();
if (i < 0 || j < 0 || i >= n || j >= n || (grid[i][j] != 0 && grid[i][j] <= dist)) return;
grid[i][j] = dist;
helper(grid, i - 1, j, dist + 1);
helper(grid, i + 1, j, dist + 1);
helper(grid, i, j - 1, dist + 1);
helper(grid, i, j + 1, dist + 1);
}
};
其實這道題的最優解法並不是 BFS 或者 DFS，而是下面這種兩次掃描的方法，在之前那道 01 Matrix 中就使用過。有點像動態規劃的感覺，還是建立距離場，先從左上遍歷到右下，遇到1的位置跳過，然後初始化0位置的值為 201（因為n不超過 100，所以距離不會超過 200），然後用左邊和上邊的值加1來更新當前位置的，注意避免越界。然後從右邊再遍歷到左上，還是遇到1的位置跳過，然後用右邊和下邊的值加1來更新當前位置的，注意避免越界，同時還要更新結果 res 的值。最終若 res 為 201，則返回 -1，否則返回 res-1，參見程式碼如下：
解法三：
class Solution {
public:
int maxDistance(vector<vector<int>>& grid) {
int res = 0, n = grid.size();
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) continue;
grid[i][j] = 201;
if (i > 0) grid[i][j] = min(grid[i][j], grid[i - 1][j] + 1);
if (j > 0) grid[i][j] = min(grid[i][j], grid[i][j - 1] + 1);
}
}
for (int i = n - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
if (grid[i][j] == 1) continue;
if (i < n - 1) grid[i][j] = min(grid[i][j], grid[i + 1][j] + 1);
if (j < n - 1) grid[i][j] = min(grid[i][j], grid[i][j + 1] + 1);
res = max(res, grid[i][j]);
}
}
return res == 201 ? -1 : res - 1;
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/1162
類似題目：
Shortest Distance from All Buildings
參考資料：
https://leetcode.com/problems/as-far-from-land-as-possible/