Problem0778

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778. Swim in Rising Water

题目

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
Explanation:
 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

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Note:

1. 2 <= N <= 50.
2. grid[i][j] is a permutation of [0, ..., N*N - 1].

解题思路

见程序注释

Documentation

Index

• type PQ
  • func (pq PQ) Len() int
  • func (pq PQ) Less(i, j int) bool
  • func (pq *PQ) Pop() interface{}
  • func (pq *PQ) Push(x interface{})
  • func (pq PQ) Swap(i, j int)

Types

type PQ

type PQ []*entry

PQ implements heap.Interface and holds entries.

func (PQ) Len

func (pq PQ) Len() int

func (PQ) Less

func (pq PQ) Less(i, j int) bool

func (*PQ) Pop

func (pq *PQ) Pop() interface{}

Pop 从 pq 中取出最优先的 entry

func (*PQ) Push

func (pq *PQ) Push(x interface{})

Push 往 pq 中放 entry

func (PQ) Swap

func (pq PQ) Swap(i, j int)