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Commit 34e9f2e

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Add C++ solutions for leetcode 695.
1 parent 5798d7b commit 34e9f2e

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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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int dx[4] = {-1, 0, 1, 0};
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int dy[4] = {0, 1, 0, -1};
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int m, n;
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public:
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int maxAreaOfIsland(vector<vector<int>>& grid) {
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m = grid.size();
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n = grid[0].size();
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int res = 0;
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for (int i = 0; i < m; i++)
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{
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for (int j = 0; j < n; j++)
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{
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if (grid[i][j] == 1)
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res = max(res, dfs(grid, i, j));
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}
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}
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return res;
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}
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int dfs(vector<vector<int>>& grid, int x, int y) /* G[i][j] == 1时才调用dfs */
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{
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int res = 1; // 把当前格子加进来
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grid[x][y] = 0; /* 加完之后就将用水(0)代替之前的陆地(1), 目的是避免重复访问。用这种方法就不需要使用visited数组记录状态了 */
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for (int i = 0; i < 4; i++)
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{
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auto a = x + dx[i];
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auto b = y + dy[i];
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if (a >= 0 && a < m && b >= 0 && b < n) // 确保没越界
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{
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if (grid[a][b] == 1)
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res += dfs(grid, a, b); // 把连通块数累加一下
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}
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}
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return res;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<vector<int>> grid =
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{
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{0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0},
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{0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},
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{0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0},
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{0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0}
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};
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auto res = sol.maxAreaOfIsland(grid);
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cout << res << endl;
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return 0;
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}
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#include<vector>
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#include<algorithm>
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#include<iostream>
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using namespace std;
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class Solution {
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int dx[4] = {-1, 0, 1, 0};
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int dy[4] = {0, 1, 0, -1};
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int m, n;
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public:
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int maxAreaOfIsland(vector<vector<int>>& grid) {
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m = grid.size();
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n = grid[0].size();
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int res = 0;
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/* 由于1 <= m, n <= 50, 故不需要判空 */
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int maxArea = 0;
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for (int i = 0; i < m; i++)
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{
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for (int j = 0; j < n; j++)
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maxArea = max(maxArea, dfs(grid, i, j));
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}
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return maxArea;
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}
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int dfs(vector<vector<int>>& grid, int x, int y)
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{
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// 判断当前传入坐标位置是否越界或是否为水域
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if(x < 0 || x >= m || y < 0 || y >= n || grid[x][y] == 0)
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return 0;
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grid[x][y] = 0; // 已访问过的岛屿也修改为0,避免重复访问
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int area = 1; // 每个岛屿的面积为1
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for (int i = 0; i < 4; i++)
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{
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auto a = x + dx[i];
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auto b = y + dy[i];
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area += dfs(grid, a, b);
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}
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return area;
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}
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};
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// Test
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int main()
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{
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Solution sol;
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vector<vector<int>> grid =
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{
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{0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0},
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{0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0},
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{0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0},
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{0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0},
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{0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0}
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};
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auto res = sol.maxAreaOfIsland(grid);
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cout << res << endl;
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return 0;
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}

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