标签：The and ges amp bsp ret turn note matrix
Given an m x n matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.
Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.
Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.
Note:
Example:
Given the following 5x5 matrix:
Pacific ~ ~ ~ ~ ~
~ 1 2 2 3 (5) *
~ 3 2 3 (4) (4) *
~ 2 4 (5) 3 1 *
~ (6) (7) 1 4 5 *
~ (5) 1 1 2 4 *
* * * * * Atlantic
Return:
[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
解法1：DFS，遍历每一个元素，找到路径，复杂度过高，有些case超时
class Solution { public: vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) { vector<vector<int>>res; int m = matrix.size(); if (m < 1)return res; int n = matrix[0].size(); if (n < 1)return res; for (int i = 0; i < m; ++i) { for(int j = 0; j < n; ++j) { bool f1 = false, f2 = false; vector<vector<bool>> marked(m, vector<bool>(n, false)); dfs(matrix, i, j, f1, f2, marked); if (f1 && f2)res.push_back(vector<int>{i,j}); } } return res; } void dfs(vector<vector<int>>& matrix, int i, int j, bool& flag1, bool& flag2, vector<vector<bool>>& marked) { if ((flag1 && flag2) || marked[i][j])return; int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; marked[i][j] = true; for (auto d: dir) { if (i+d[0] < 0 || j+d[1] < 0) // 当前格子已经第一排或第一列了 { flag1 = true; continue; } if (i+d[0] >= matrix.size() || j+d[1] >= matrix[0].size()) // 当前格子已经最后一排或者最后一列了 { flag2 = true; continue; } // 下一个方格比当前的低，并且之前没有走过 if(matrix[i+d[0]][j+d[1]] <= matrix[i][j] && !marked[i+d[0]][j+d[1]]) dfs(matrix, i+d[0], j+d[1], flag1, flag2, marked); if (flag1 && flag2)return; } } };
解法2：DFS从太平洋和大西洋往内部遍历
class Solution { public: vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) { vector<vector<int>>res; int m = matrix.size(); if (m < 1)return res; int n = matrix[0].size(); if (n < 1)return res; vector<vector<bool>> marked1(m, vector<bool>(n, false)); vector<vector<bool>> marked2(m, vector<bool>(n, false)); for (int i = 0;i < m; ++i) { dfs(matrix, i, 0, marked1); dfs(matrix, i, n-1, marked2); } for (int i = 0;i < n;++i) { dfs(matrix, 0, i, marked1); dfs(matrix, m-1, i, marked2); } for (int i = 0; i < m; ++i) { for(int j = 0; j < n; ++j) { if (marked1[i][j] && marked2[i][j])res.push_back(vector<int>{i,j}); } } return res; } void dfs(vector<vector<int>>& matrix, int i, int j, vector<vector<bool>>& marked) { if (marked[i][j])return; int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}}; marked[i][j] = true; for (auto d: dir) { int i_next = i + d[0]; int j_next = j + d[1]; if(i_next < 0 || i_next >= matrix.size() || j_next < 0 || j_next >= matrix[0].size() || matrix[i][j] > matrix[i_next][j_next]) continue; dfs(matrix, i+d[0], j+d[1], marked); } } };
417. Pacific Atlantic Water Flow
标签：The and ges amp bsp ret turn note matrix
原文地址：https://www.cnblogs.com/qiang-wei/p/12359484.html
