[LeetCode] Walls and Gates 牆和門 LeetCode All in One 題目講解彙總(持續更新中...)

時間 2019-12-08

標籤 leetcode walls gates 題目講解彙總持續更新简体版

原文原文鏈接

You are given a m x n 2D grid initialized with these three possible values.html

-1 - A wall or an obstacle.
0 - A gate.
INF - Infinity means an empty room. We use the value 2³¹ - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.java

For example, given the 2D grid:
less

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF

After running your function, the 2D grid should be:
post

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

這道題相似一種迷宮問題，規定了-1表示牆，0表示門，讓求每一個點到門的最近的曼哈頓距離，這其實相似於求距離場Distance Map的問題，那麼咱們先考慮用DFS來解，思路是，咱們搜索0的位置，每找到一個0，以其周圍四個相鄰點爲起點，開始DFS遍歷，並帶入深度值1，若是遇到的值大於當前深度值，咱們將位置值賦爲當前深度值，並對當前點的四個相鄰點開始DFS遍歷，注意此時深度值須要加1，這樣遍歷完成後，全部的位置就被正確地更新了，參見代碼以下：ui

解法一：url

class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        for (int i = 0; i < rooms.size(); ++i) {
            for (int j = 0; j < rooms[i].size(); ++j) {
                if (rooms[i][j] == 0) dfs(rooms, i, j, 0);
            }
        }
    }
    void dfs(vector<vector<int>>& rooms, int i, int j, int val) {
        if (i < 0 || i >= rooms.size() || j < 0 || j >= rooms[i].size() || rooms[i][j] < val) return;
        rooms[i][j] = val;
        dfs(rooms, i + 1, j, val + 1);
        dfs(rooms, i - 1, j, val + 1);
        dfs(rooms, i, j + 1, val + 1);
        dfs(rooms, i, j - 1, val + 1);
    }
};

那麼下面咱們再來看BFS的解法，須要藉助queue，咱們首先把門的位置都排入queue中，而後開始循環，對於門位置的四個相鄰點，咱們判斷其是否在矩陣範圍內，而且位置值是否大於上一位置的值加1，若是知足這些條件，咱們將當前位置賦爲上一位置加1，並將次位置排入queue中，這樣等queue中的元素遍歷完了，全部位置的值就被正確地更新了，參見代碼以下：spa

解法二：rest

class Solution {
public:
    void wallsAndGates(vector<vector<int>>& rooms) {
        queue<pair<int, int>> q;
        vector<vector<int>> dirs{{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
        for (int i = 0; i < rooms.size(); ++i) {
            for (int j = 0; j < rooms[i].size(); ++j) {
                if (rooms[i][j] == 0) q.push({i, j});   
            }
        }
        while (!q.empty()) {
            int i = q.front().first, j = q.front().second; q.pop();
            for (int k = 0; k < dirs.size(); ++k) {
                int x = i + dirs[k][0], y = j + dirs[k][1];
                if (x < 0 || x >= rooms.size() || y < 0 || y >= rooms[0].size() || rooms[x][y] < rooms[i][j] + 1) continue;
                rooms[x][y] = rooms[i][j] + 1;
                q.push({x, y});
            }
        }
    }
};