标签：south ret mono col output integer nat mat jpg
Given a matrix of integers A with R rows and C columns, find the maximum score of a path starting at [0,0] and ending at [R-1,C-1].
The score of a path is the minimum value in that path. For example, the value of the path 8 → 4 → 5 → 9 is 4.
A path moves some number of times from one visited cell to any neighbouring unvisited cell in one of the 4 cardinal directions (north, east, west, south).
Example 1:
Input: [[5,4,5],[1,2,6],[7,4,6]]
Output: 4
Explanation:
The path with the maximum score is highlighted in yellow.
Example 2:
Input: [[2,2,1,2,2,2],[1,2,2,2,1,2]] Output: 2
Example 3:
Input: [[3,4,6,3,4],[0,2,1,1,7],[8,8,3,2,7],[3,2,4,9,8],[4,1,2,0,0],[4,6,5,4,3]] Output: 3
Note:
1 <= R, C <= 1000 <= A[i][j] <= 10^9Time: O(M * N * log(M * N))
class Solution { int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; int ROW; int COL; public int maximumMinimumPath(int[][] A) { ROW = A.length; COL = A[0].length; boolean[][] visited = new boolean[ROW][COL]; int minValue = A[0][0]; // find the max value for the nearby cells PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[2] - a[2]); pq.offer(new int[]{0, 0, A[0][0]}); while (!pq.isEmpty()) { int[] cur = pq.poll(); int curRow = cur[0]; int curCol = cur[1]; minValue = Math.min(minValue, cur[2]); if (curRow == ROW - 1 && curCol == COL - 1) { break; } for (int[] dir: directions) { int nextRow = curRow + dir[0]; int nextCol = curCol + dir[1]; if (nextRow < 0 || nextRow >= ROW || nextCol < 0 || nextCol >= COL || visited[nextRow][nextCol]) { continue; } pq.offer(new int[]{nextRow, nextCol, A[nextRow][nextCol]}); visited[nextRow][nextCol] = true; } } return minValue; } }
[LC] 1102. Path With Maximum Minimum Value
标签：south ret mono col output integer nat mat jpg
原文地址：https://www.cnblogs.com/xuanlu/p/12784319.html
