Open azl397985856 opened 9 months ago
class Solution:
def swimInWater(self, grid: List[List[int]]) -> int:
res = 0
n = len(grid)
heap = [(grid[0][0],0,0)]
visited = set([(0,0)])
while heap:
height,x,y = heapq.heappop(heap)
res = max(res,height)
if x == n-1 and y == n-1:
return res
for dx,dy in [(0,1),(0,-1),(1,0),(-1,0)]:
new_x,new_y = x + dx,y + dy
if 0 <= new_x < n and 0 <= new_y < n and (new_x,new_y) not in visited:
visited.add((new_x,new_y))
heapq.heappush(heap,(grid[new_x][new_y],new_x,new_y))
return -1
class Solution {
public:
int swimInWater(vector<vector<int>>& grid) {
int m=grid.size();
int n=grid[0].size();
UnionFind uf(m*n);
vector<tuple<int,int,int>>edge;
for(int i=0;i<m;i++)
for(int j=0;j<n;j++){
int id=i*n+j;
if(i>0)edge.emplace_back(max(grid[i][j],grid[i-1][j]),id,id-m);
if(j>0)edge.emplace_back(max(grid[i][j],grid[i][j-1]),id,id-1);
}
sort(edge.begin(),edge.end());
int res=0;
for(auto&[v,x,y]:edge){
uf.merge(x,y);
if(uf.connected(0,n*n-1)){
res=v;
break;
}
}
return res;
}
};
确实很难想到是二分法. 根据题解, 因为是有限解空间, 所以可以二分法. 首先就要先找到解空间,上界和下界.然后运用dfs来判断
class Solution {
public boolean dfs(int mid, int x, int y, boolean[][] visited, int[][] grid){
if (x > grid.length - 1 || x < 0 || y > grid[0].length - 1 || y < 0) return false;
if (grid[x][y] > mid) return false;
if(x == grid.length - 1&& y == grid[0].length - 1) return true;
if (visited[x][y]) return false;
visited[x][y] = true;
return dfs(mid, x + 1, y, visited, grid) || dfs(mid, x - 1, y, visited, grid) || dfs(mid, x, y + 1, visited, grid) || dfs(mid, x, y - 1, visited, grid);
}
public int swimInWater(int[][] grid) {
int l = 0;
int r = Integer.MAX_VALUE;
//先找出解空间
for(int i = 0; i< grid.length; i++){
for(int j = 0; j<grid[0].length; j++){
if(r<grid[i][j]){
r = grid[i][j];
}
}
}
// 二分法
while (l <= r) {
boolean[][] visited = new boolean[grid.length][grid[0].length];
int mid = l + (r - l) / 2;
if (dfs(mid, 0, 0, visited, grid)) {
r = mid - 1; // 成为一个备胎, 但不一定是最小的
} else {
l = mid + 1; // head towards 最小的选项
}
}
return l;
}
}
class Solution:
def swimInWater(self, grid: List[List[int]]) -> int:
res = 0
n = len(grid)
heap = [(grid[0][0],0,0)]
visited = set([(0,0)])
while heap:
height,x,y = heapq.heappop(heap)
res = max(res,height)
if x == n-1 and y == n-1:
return res
for dx,dy in [(0,1),(0,-1),(1,0),(-1,0)]:
new_x,new_y = x + dx,y + dy
if 0 <= new_x < n and 0 <= new_y < n and (new_x,new_y) not in visited:
visited.add((new_x,new_y))
heapq.heappush(heap,(grid[new_x][new_y],new_x,new_y))
return -1
二分查找+深度优先遍历
在方格数值区间[0,N*N-1]中找一个数作为等待时间,使其满足:存在从左上角到右下角的路径,且该数要最小
public class Solution {
private int N;
public int[][] DIRECTIONS = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
public int swimInWater(int[][] grid) {
this.N = grid.length;
int left = 0;
int right = N * N - 1;
while (left < right) {
int mid = (left + right) / 2;
boolean[][] visited = new boolean[N][N];
if (grid[0][0] <= mid && dfs(grid, 0, 0, visited, mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
private boolean dfs(int[][] grid, int x, int y, boolean[][] visited, int threshold) {
visited[x][y] = true;
for (int[] direction : DIRECTIONS) {
int newX = x + direction[0];
int newY = y + direction[1];
if (inArea(newX, newY) && !visited[newX][newY] && grid[newX][newY] <= threshold) {
if (newX == N - 1 && newY == N - 1) {
return true;
}
if (dfs(grid, newX, newY, visited, threshold)) {
return true;
}
}
}
return false;
}
private boolean inArea(int x, int y) {
return x >= 0 && x < N && y >= 0 && y < N;
}
}
var swimInWater = function (grid) { let n = grid.length let low = grid[0][0] let high = n * n while (low < high) { let mid = Math.floor((low + high) / 2) if (!isCanSwim(mid, grid)) { low = mid + 1 } else { high = mid } } return low }; const isCanSwim = (t, grid) => { let n = grid.length let vis = Array.from(Array(n), () => Array(n).fill(false)) let dfs = (row, col) => { if (row < 0 || col < 0 || row >= n || col >= n || vis[row][col] || grid[row][col] > t) { return false } if (row == n - 1 && col == n - 1) { return true } vis[row][col] = true return dfs(row + 1, col) || dfs(row - 1, col) || dfs(row, col + 1) || dfs(row, col - 1); } return dfs(0, 0) }
778. 水位上升的泳池中游泳
入选理由
暂无
题目地址
https://leetcode-cn.com/problems/swim-in-rising-water
前置知识
暂无
题目描述
在一个 N x N 的坐标方格 grid 中,每一个方格的值 grid[i][j] 表示在位置 (i,j) 的平台高度。
现在开始下雨了。当时间为 t 时,此时雨水导致水池中任意位置的水位为 t 。你可以从一个平台游向四周相邻的任意一个平台,但是前提是此时水位必须同时淹没这两个平台。假定你可以瞬间移动无限距离,也就是默认在方格内部游动是不耗时的。当然,在你游泳的时候你必须待在坐标方格里面。
你从坐标方格的左上平台 (0,0) 出发。最少耗时多久你才能到达坐标方格的右下平台 (N-1, N-1)?
示例 1:
输入: [[0,2],[1,3]] 输出: 3 解释: 时间为 0 时,你位于坐标方格的位置为 (0, 0)。 此时你不能游向任意方向,因为四个相邻方向平台的高度都大于当前时间为 0 时的水位。
等时间到达 3 时,你才可以游向平台 (1, 1). 因为此时的水位是 3,坐标方格中的平台没有比水位 3 更高的,所以你可以游向坐标方格中的任意位置 示例 2:
输入: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]] 输出: 16 解释: 0 1 2 3 4 24 23 22 21 5 12 13 14 15 16 11 17 18 19 20 10 9 8 7 6
最终的路线用加粗进行了标记。 我们必须等到时间为 16,此时才能保证平台 (0, 0) 和 (4, 4) 是连通的
提示:
2 <= N <= 50. grid[i][j] 位于区间 [0, ..., N*N - 1] 内。