http://www.itdadao.com/articles/c15a774486p0.htmlzhttps://leetcode.com/problems/island-perimeter/
这道题给了我们一个格子图，若干连在一起的格子形成了一个小岛，规定了图中只有一个相连的岛，且岛中没有湖，让我们求岛的周长。我们知道一个格子有四条边，但是当两个格子相邻，周围为6，若某个格子四周都有格子，那么这个格子一条边都不算在周长里。那么我们怎么统计出岛的周长呢？第一种方法，我们对于每个格子的四条边分别来处理，首先看左边的边，只有当左边的边处于第一个位置或者当前格子的左面没有岛格子的时候，左边的边计入周长。其他三条边的分析情况都跟左边的边相似
http://blog.csdn.net/mebiuw/article/details/53265123
public int islandPerimeter(int[][] grid) { int permeter = 0; int n = grid.length; int m = grid[0].length; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if(grid[i][j] == 1){ if(i==0 || grid[i-1][j] == 0) permeter++; if(i==n-1 || grid[i+1][j] == 0) permeter++; if(j==0 || grid[i][j-1] == 0) permeter++; if(j==m-1 || grid[i][j+1] == 0) permeter++; } } } return permeter; }
http://blog.csdn.net/jmspan/article/details/53254264
方法：逐个扫描。
X.
下面这种方法对于每个岛屿格子先默认加上四条边，然后检查其左面和上面是否有岛屿格子，有的话分别减去两条边，这样也能得到正确的结果
http://bookshadow.com/weblog/2016/11/20/leetcode-island-perimeter/
X.
https://discuss.leetcode.com/topic/68786/clear-and-easy-java-solution
https://discuss.leetcode.com/topic/68751/easy-dfs-solution-explaination-without-visited-array
You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
Example:
[[0,1,0,0], [1,1,1,0], [0,1,0,0], [1,1,0,0]] Answer: 16 Explanation: The perimeter is the 16 yellow stripes in the image below:https://www.cnblogs.com/grandyang/p/6096138.html
这道题给了我们一个格子图，若干连在一起的格子形成了一个小岛，规定了图中只有一个相连的岛，且岛中没有湖，让我们求岛的周长。我们知道一个格子有四条边，但是当两个格子相邻，周围为6，若某个格子四周都有格子，那么这个格子一条边都不算在周长里。那么我们怎么统计出岛的周长呢？第一种方法，我们对于每个格子的四条边分别来处理，首先看左边的边，只有当左边的边处于第一个位置或者当前格子的左面没有岛格子的时候，左边的边计入周长。其他三条边的分析情况都跟左边的边相似
http://blog.csdn.net/mebiuw/article/details/53265123
public int islandPerimeter(int[][] grid) { int permeter = 0; int n = grid.length; int m = grid[0].length; for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if(grid[i][j] == 1){ if(i==0 || grid[i-1][j] == 0) permeter++; if(i==n-1 || grid[i+1][j] == 0) permeter++; if(j==0 || grid[i][j-1] == 0) permeter++; if(j==m-1 || grid[i][j+1] == 0) permeter++; } } } return permeter; }
http://blog.csdn.net/jmspan/article/details/53254264
方法：逐个扫描。
- public int islandPerimeter(int[][] grid) {
- int p = 0;
- for(int i = 0; i < grid.length; i++) {
- for(int j = 0; j < grid[i].length; j++) {
- if (grid[i][j] == 0) continue;
- if (i == 0 || grid[i - 1][j] == 0) p++;
- if (i == grid.length - 1 || grid[i + 1][j] == 0) p++;
- if (j == 0 || grid[i][j - 1] == 0) p++;
- if (j == grid[i].length - 1 || grid[i][j + 1] == 0) p++;
- }
- }
- return p;
- }
X.
下面这种方法对于每个岛屿格子先默认加上四条边，然后检查其左面和上面是否有岛屿格子，有的话分别减去两条边，这样也能得到正确的结果
int islandPerimeter(vector<vector<int>>& grid) { if (grid.empty() || grid[0].empty()) return 0; int res = 0, m = grid.size(), n = grid[0].size(); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid[i][j] == 0) continue; res += 4; if (i > 0 && grid[i - 1][j] == 1) res -= 2; if (j > 0 && grid[i][j - 1] == 1) res -= 2; } } return res; }
每一个陆地单元格的周长为4，当两单元格上下或者左右相邻时，令周长减2。
https://leetcode.com/problems/island-perimeter/discuss/94992/java-9-line-solution-add-4-for-each-land-and-remove-2-for-each-internal-edgepublic static int islandPerimeter(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) return 0;
int result = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
result += 4;
if (i > 0 && grid[i-1][j] == 1) result -= 2;
if (j > 0 && grid[i][j-1] == 1) result -= 2;
}
}
}
return result;
}X.
https://discuss.leetcode.com/topic/68786/clear-and-easy-java-solution
- loop over the matrix and count the number of islands;
- if the current dot is an island, count if it has any right neighbour or down neighbour;
- the result is islands * 4 - neighbours * 2
public int islandPerimeter(int[][] grid) {
int islands = 0, neighbours = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[i].length; j++) {
if (grid[i][j] == 1) {
islands++; // count islands
if (i < grid.length - 1 && grid[i + 1][j] == 1) neighbours++; // count down neighbours
if (j < grid[i].length - 1 && grid[i][j + 1] == 1) neighbours++; // count right neighbours
}
}
}
return islands * 4 - neighbours * 2;
}https://discuss.leetcode.com/topic/68751/easy-dfs-solution-explaination-without-visited-array
he idea here is that each land cell contributes as many lines in perimeter as it's surrounded by water / boundary.
void dfs(vector<vector<int>>& b, int *ans, int i, int j) {
if (i < 0 || i >= b.size() || j < 0 || j >= b[0].size() || b[i][j] != 1)
return;
b[i][j] = -1; // mark it as visited
*ans += (j + 1 >= b[0].size() || b[i][j+1] == 0) /* right */ +
(i - 1 < 0 || b[i-1][j] == 0) /* top */ +
(j - 1 < 0 || b[i][j-1] == 0) /* left */ +
(i + 1 >= b.size() || b[i+1][j] == 0) /* bottom */;
dfs(b, ans, i, j + 1);
dfs(b, ans, i - 1, j);
dfs(b, ans, i, j - 1);
dfs(b, ans, i + 1, j);
return;
}
int islandPerimeter(vector<vector<int>>& grid) {
int ans = 0, i, j;
for (i = 0; i < grid.size(); i++) {
for (j = 0; j < grid[0].size(); j++) {
if (grid[i][j]) {
dfs(grid, &ans, i, j);
return ans;
}
}
}
return 0;
}
解法：如果只有一个点，那么周长就是4，如果这个点和其他的一个点相邻，那么对于周长的贡献就要少一。同理，对于另外一个相邻的点，那么
贡献的周长也少一。于是先遍历，看看有多少个为1的点，算出“最大”的周长以后，减去因为相邻而减少的周长长度就可以了。
贡献的周长也少一。于是先遍历，看看有多少个为1的点，算出“最大”的周长以后，减去因为相邻而减少的周长长度就可以了。
public int islandPerimeter(int[][] grid) {
boolean[][] mark = new boolean[grid.length][grid[0].length];
int count = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
mark[i][j] = true;
count++;
} else {
mark[i][j] = false;
}
}
}
int perimeter = count * 4;
System.out.println("perimeter:" + perimeter);
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (mark[i][j]) {
perimeter -= neigbors(i, j, mark); //减去相邻的点
}
}
}
return perimeter;
}
private int neigbors(int i, int j, boolean[][] mark) {
int count = 0;
for (int x = -1; x <= 1; x += 2) {
int tempx = x + i;
int tempy = j;
if (isSafe(tempx, tempy, mark.length, mark[0].length) && mark[tempx][tempy]) {
count++;
}
}
for (int y = -1; y <= 1; y += 2) {
int tempx = i;
int tempy = y + j;
if (isSafe(tempx, tempy, mark.length, mark[0].length) && mark[tempx][tempy]) {
count++;
}
}
System.out.printf("i:%d,j:%d\n", i, j);
System.out.println("count:" + count);
return count;
}
private boolean isSafe(int x, int y, int xlength, int ylength) {
if (x < 0 || x >= xlength || y < 0 || y >= ylength) {
return false;
} else {
return true;
}
}
http://www.geeksforgeeks.org/find-perimeter-shapes-formed-1s-binary-matrix/
The idea is to traverse the matrix, find all ones and find their contribution in perimeter. The maximum contribution of a 1 is four if it is surrounded by all 0s. The contribution reduces by one with 1 around it.
Algorithm for solving this problem:
- Traverse the whole matrix and find the cell having value equal to 1.
- Calculate the number of closed side for that cell and add, 4 – number of closed side to the total perimeter.
int numofneighbour(int mat[][C], int i, int j){ int count = 0; // UP if (i > 0 && mat[i - 1][j]) count++; // LEFT if (j > 0 && mat[i][j - 1]) count++; // DOWN if (i < R-1 && mat[i + 1][j]) count++; // RIGHT if (j < C-1 && mat[i][j + 1]) count++; return count;}// Returns sum of perimeter of shapes formed with 1sint findperimeter(int mat[R][C]){ int perimeter = 0; // Traversing the matrix and finding ones to // calculate their contribution. for (int i = 0; i < R; i++) for (int j = 0; j < C; j++) if (mat[i][j]) perimeter += (4 - numofneighbour(mat, i ,j)); return perimeter;}