Add C++ solutions for leetcode 1162.

Author: yanglr

Diff for: cpp-leetcode/leetcode1162-as-far-from-land-as-possible_bfs1.cpp

@@ -0,0 +1,68 @@
+#include<vector>
+#include<queue>
+#include<algorithm>
+#include<iostream>
+using namespace std;
+
+class Solution {
+    const int dx[4] = {0, 1, 0, -1};
+    const int dy[4] = {1, 0, -1, 0};    
+public:
+    int maxDistance(vector<vector<int>>& grid) {
+        int N = grid.size();
+        queue<pair<int, int>> q; // queue存储坐标值, 即 pair of {x, y}
+        vector<vector<int>> d(N, vector<int>(N, 1000)); // 二维数组d[][]: 记录每个格子grid[i][j]的距离值
+        for (int i = 0; i < N; i++)
+        {
+            for (int j = 0; j < N; j++)
+                if (grid[i][j] == 1)
+                {
+                    q.push(make_pair(i, j));
+                    d[i][j] = 0;
+                }            
+        }
+        while (!q.empty())
+        {
+            auto kvp = q.front();
+            q.pop();
+            for (int i = 0; i < 4; i++)
+            {
+                int newX = kvp.first + dx[i], newY = kvp.second + dy[i];
+                if (newX < 0 || newX >= N || newY < 0 || newY >= N) // 越界了, 跳过
+                    continue;
+
+                if (d[newX][newY] > d[kvp.first][kvp.second] + 1) /* 如果从水域(值为0的格子)走到陆地(值为1的格子)或从陆地走到水域, 且新到达的位置之前没被访问过, 此时需要将其坐标加入队列, 并更新距离值d */
+                {
+                    d[newX][newY] = d[kvp.first][kvp.second] + 1; /* 当前格子的上下左右4个方向之一走一步恰好使得曼哈顿距离增加1 */
+                    q.push(make_pair(newX, newY));
+                }
+            }
+        }
+        int res = -1;
+        for (int i = 0; i < N; i++)
+        {
+            for (int j = 0; j < N; j++)
+            {
+                if (grid[i][j] == 0 && d[i][j] <= 200) /* 挑出访问过的水域位置(值为0的格子), 并维护这些格子中距离值d的最大值 */
+                    res = max(res, d[i][j]);
+            }
+        }
+        return res;
+    }
+};
+
+// Test
+int main()
+{
+    Solution sol;
+    vector<vector<int>> grid = 
+    {
+        {1, 0, 1}, 
+        {0, 0, 0}, 
+        {1, 0, 1}
+    };
+    auto res = sol.maxDistance(grid);
+    cout << res << endl;
+
+    return 0;
+}

Diff for: cpp-leetcode/leetcode1162-as-far-from-land-as-possible_bfs2.cpp

@@ -0,0 +1,70 @@
+#include<vector>
+#include<queue>
+#include<algorithm>
+#include<iostream>
+using namespace std;
+
+class Solution {
+    const int dx[4] = {0, 1, 0, -1};
+    const int dy[4] = {1, 0, -1, 0};    
+public:
+    int maxDistance(vector<vector<int>>& grid) {
+        const int N = grid.size();
+        deque<pair<int, int>> q;
+        int d[N][N];
+        for (auto& row : d)
+            fill(row, row + N, 1000);
+        for (int i = 0; i < N; i++)
+        {
+            for (int j = 0; j < N; j++)
+                if (grid[i][j] == 1)
+                {
+                    q.push_back(make_pair(i, j));
+                    d[i][j] = 0;
+                }            
+        }
+        while (!q.empty())
+        {
+            auto kvp = q.front();
+            q.pop_front();
+            for (int i = 0; i < 4; i++)
+            {
+                int newX = kvp.first + dx[i], newY = kvp.second + dy[i];
+                if (newX < 0 || newX >= N || newY < 0 || newY >= N)
+                    continue;
+
+                if (d[newX][newY] > d[kvp.first][kvp.second] + 1)
+                {
+                    d[newX][newY] = d[kvp.first][kvp.second] + 1; 
+                    q.push_back(make_pair(newX, newY));
+                }
+            }
+        }
+        int res = -1;
+        for (int i = 0; i < N; i++)
+        {
+            for (int j = 0; j < N; j++)
+            {
+                if (grid[i][j] == 0 && d[i][j] <= 200)
+                    res = max(res, d[i][j]);
+            }
+        }
+        return res;
+    }
+};
+
+// Test
+int main()
+{
+    Solution sol;
+    vector<vector<int>> grid = 
+    {
+        {1, 0, 1}, 
+        {0, 0, 0}, 
+        {1, 0, 1}
+    };
+    auto res = sol.maxDistance(grid);
+    cout << res << endl;
+
+    return 0;
+}