LWC 70: 778. Swim in Rising Water
LWC 70: 778. Swim in Rising Water
LWC 70: 778. Swim in Rising Water
Problem:
On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j). Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim. You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?
Example 1:
Input: [[0,2],[1,3]] Output: 3 Explanation: At time 0, you are in grid location (0, 0). You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0. You cannot reach point (1, 1) until time 3. When the depth of water is 3, we can swim anywhere inside the grid.
Example 2:
Input: [[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]] Output: 16 The final route is marked in bold. We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
Note:
- 2 <= N <= 50.
- grid[i][j] is a permutation of [0, …, N*N - 1].
思路：
N很小，只有50，可以暴力解决。对于每一个时刻t，grid变成max(t, grid[i][j])，对grid更新后，所有<t的值都变成了t，在这些t值上从(0， 0)dfs至(n - 1, n - 1)，如果能够找到可行的路径，输出t即可。
代码如下（125ms）：
public int swimInWater(int[][] grid) {
N = grid.length;
for (int i = grid[0][0]; i <= N * N; ++i) {
if (go(grid, 0, 0, new boolean[N][N], i)) {
return i;
}
}
return -1;
}
int[][] dir = {{-1, 0},{0, 1},{0, -1},{1, 0}};
int N = 0;
boolean go(int[][] grid, int i, int j, boolean[][] vis, int mid) {
if (i == N - 1 && j == N - 1) return true;
vis[i][j] = true;
for (int[] d : dir) {
int ni = i + d[0];
int nj = j + d[1];
if (ni >= 0 && ni < N && nj >= 0 && nj < N && !vis[ni][nj] && Math.max(grid[ni][nj], mid) == Math.max(grid[i][j], mid)) {
if (go(grid, ni, nj, vis, mid)) return true;
}
}
return false;
}不够快，来个二分版本（16ms）：
public int swimInWater(int[][] grid) {
N = grid.length;
int lf = 0;
int rt = N * N;
while (lf < rt) {
int mid = lf + (rt - lf) / 2;
if (go(grid, 0, 0, new boolean[N][N], mid)) {
rt = mid;
}
else {
lf = mid + 1;
}
}
return rt;
}
int[][] dir = {{-1, 0},{0, 1},{0, -1},{1, 0}};
int N = 0;
boolean go(int[][] grid, int i, int j, boolean[][] vis, int mid) {
if (i == N - 1 && j == N - 1) return true;
vis[i][j] = true;
for (int[] d : dir) {
int ni = i + d[0];
int nj = j + d[1];
if (ni >= 0 && ni < N && nj >= 0 && nj < N && !vis[ni][nj] && Math.max(grid[ni][nj], mid) == Math.max(grid[i][j], mid)) {
if (go(grid, ni, nj, vis, mid)) return true;
}
}
return false;
}Python版本：
class Solution(object):
def swimInWater(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
N = len(grid)
def go(i, j, mid, vis):
if i == N - 1 and j == N - 1: return True
vis.add((i, j))
for d in (-1, 1):
for ni, nj in [(i + d, j), (i, j + d)]:
if 0 <= ni < N and 0 <= nj < N and (ni, nj) not in vis and max(mid, grid[i][j]) == max(mid, grid[ni][nj]):
if go(ni, nj, mid, vis): return True
return False
lf = 0
rt = N * N
while lf < rt:
mid = (lf + rt) // 2
if go(0, 0, mid, set()):
rt = mid
else:
lf = mid + 1
return rt腾讯云开发者
