417. 太平洋大西洋水流问题

代码使用队列进行广度搜索，分别遍历太平洋 和大西洋的河流，取交集。

class Solution {
public:vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {set<pair<int,int>> pacSet;set<pair<int,int>> atlSet;vector<vector<int>> ans;queue<pair<int,int>> que; int m = heights.size();int n = heights[0].size();for(int i=0;i<n;++i){que.push({0,i});pacSet.insert({0,i});}for(int i=1;i<m;++i){que.push({i,0});pacSet.insert({i,0});}int dx[4] = {1,0,-1,0};int dy[4] = {0,1,0,-1};while(!que.empty()){auto t = que.front();que.pop();int x = t.first;int y = t.second;for(int i=0;i<4;++i){int nx = x + dx[i];int ny = y + dy[i];if(nx>=0 && nx<m && ny>=0 && ny <n && !pacSet.count({nx,ny}) && heights[nx][ny] >= heights[x][y]){que.push({nx,ny});pacSet.insert({nx,ny});}}}for(int i=0;i<n;++i){que.push({m-1,i});atlSet.insert({m-1,i});}for(int i=0;i<m-1;++i){que.push({i,n-1});atlSet.insert({i,n-1});}while(!que.empty()){auto t = que.front();que.pop();int x = t.first;int y = t.second;if(pacSet.count({x,y})){ans.push_back({x,y});}for(int i=0;i<4;++i){int nx = x + dx[i];int ny = y + dy[i];if(nx>=0 && nx<m && ny>=0 && ny <n && !atlSet.count({nx,ny}) && heights[nx][ny] >= heights[x][y]){que.push({nx,ny});atlSet.insert({nx,ny});}}}return  ans;}
};