/**
* Return an array of arrays of size *returnSize.
* The sizes of the arrays are returned as *returnColumnSizes array.
* Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
*/
int m,n;
int axis[5]={-1,0,1,0,-1};
void dfs(int** matrix,bool reach_sea[m][n],int x,int y)
{
if(reach_sea[x][y])//判断边界,终止条件
return;
reach_sea[x][y]=true;
for(int i=0;i<4;i++)//遍历每一种情况
{
int new_x=axis[i]+x;
int new_y=axis[i+1]+y;
if(new_x>=0&&new_x<m&&new_y>=0&&new_y<n&&matrix[x][y]<=matrix[new_x][new_y])//check满足
dfs(matrix,reach_sea,new_x,new_y);//继续搜索
}
}
int** pacificAtlantic(int** matrix, int matrixSize, int* matrixColSize, int* returnSize, int** returnColumnSizes){
int** res=(int**)malloc(sizeof(int*)*3000);
*returnSize=0;
if(matrixSize<1||matrixColSize[0]<1)
return res;
m=matrixSize;
n=matrixColSize[0];
*returnColumnSizes=(int*)malloc(sizeof(int)*3000);
bool reach_p[m][n];
bool reach_a[m][n];
memset(reach_a,false,sizeof(reach_a));
memset(reach_p,false,sizeof(reach_p));
for(int i=0;i<m;i++)//从左边和右边开始dfs
{
dfs(matrix,reach_a,i,0);
dfs(matrix,reach_p,i,n-1);
}
for(int i=0;i<n;i++)//从上边和下边开始dfs
{
dfs(matrix,reach_a,0,i);
dfs(matrix,reach_p,m-1,i);
}
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
{
if(reach_p[i][j]&&reach_a[i][j])//如果两个海洋都可到达,加入结果数组中
{
res[*returnSize]=(int*)malloc(sizeof(int)*2);
res[*returnSize][0]=i;
res[*returnSize][1]=j;
(*returnColumnSizes)[(*returnSize)++]=2;
}
}
return res;
}
力扣417. 太平洋大西洋水流问题
最新推荐文章于 2025-05-15 11:11:18 发布