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170 lines
6.1 KiB
Markdown
170 lines
6.1 KiB
Markdown
# [1631. Path With Minimum Effort](https://leetcode.com/problems/path-with-minimum-effort/)
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## 题目
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You are a hiker preparing for an upcoming hike. You are given `heights`, a 2D array of size `rows x columns`, where `heights[row][col]` represents the height of cell `(row, col)`. You are situated in the top-left cell, `(0, 0)`, and you hope to travel to the bottom-right cell, `(rows-1, columns-1)` (i.e., **0-indexed**). You can move **up**, **down**, **left**, or **right**, and you wish to find a route that requires the minimum **effort**.
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A route's **effort** is the **maximum absolute difference** in heights between two consecutive cells of the route.
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Return *the minimum **effort** required to travel from the top-left cell to the bottom-right cell.*
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**Example 1:**
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```
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Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
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Output: 2
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Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
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This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
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```
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**Example 2:**
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```
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Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
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Output: 1
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Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].
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```
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**Example 3:**
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```
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Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
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Output: 0
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Explanation: This route does not require any effort.
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```
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**Constraints:**
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- `rows == heights.length`
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- `columns == heights[i].length`
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- `1 <= rows, columns <= 100`
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- `1 <= heights[i][j] <= 10^6`
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## 题目大意
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你准备参加一场远足活动。给你一个二维 `rows x columns` 的地图 `heights` ，其中 `heights[row][col]` 表示格子 `(row, col)` 的高度。一开始你在最左上角的格子 `(0, 0)` ，且你希望去最右下角的格子 `(rows-1, columns-1)` （注意下标从 0 开始编号）。你每次可以往 上，下，左，右 四个方向之一移动，你想要找到耗费 体力 最小的一条路径。一条路径耗费的 体力值 是路径上相邻格子之间 高度差绝对值 的 最大值 决定的。请你返回从左上角走到右下角的最小 体力消耗值 。
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## 解题思路
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- 此题和第 778 题解题思路完全一致。在第 778 题中求的是最短连通时间。此题求的是连通路径下的最小体力值。都是求的最小值，只是 2 个值的意义不同罢了。
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- 按照第 778 题的思路，本题也有多种解法。第一种解法是 DFS + 二分。先将题目变换一个等价问法。题目要求找到最小体力消耗值，也相当于问是否存在一个体力消耗值 x，只要大于等于 x，一定能连通。利用二分搜索来找到这个临界值。体力消耗值是有序的，此处满足二分搜索的条件。题目给定柱子高度是 [1,10^6]，所以体力值一定在 [0,10^6-1] 这个区间内。判断是否取中值的条件是用 DFS 或者 BFS 搜索 (0,0) 点和 (N-1, N-1) 点之间是否连通。时间复杂度：O(mnlogC)，其中 m 和 n 分别是地图的行数和列数，C 是格子的最大高度。C 最大为 10^6，所以 logC 常数也很小。空间复杂度 O(mn)。
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- 第二种解法是并查集。将图中所有边按照权值从小到大进行排序，并依次加入并查集中。直到加入一条权值为 x 的边以后，左上角到右下角连通了。最小体力消耗值也就找到了。注意加入边的时候，只加入 `i-1` 和 `i` ，`j-1` 和 `j` 这 2 类相邻的边。因为最小体力消耗意味着不走回头路。上下左右四个方向到达一个节点，只可能从上边和左边走过来。从下边和右边走过来肯定是浪费体力了。时间复杂度：O(mnlog(mn))，其中 m 和 n 分别是地图的行数和列数，图中的边数为 O(mn)。空间复杂度 O(mn)，即为存储所有边以及并查集需要的空间。
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## 代码
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```go
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package leetcode
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import (
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"sort"
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"github.com/halfrost/LeetCode-Go/template"
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)
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var dir = [4][2]int{
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{0, 1},
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{1, 0},
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{0, -1},
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{-1, 0},
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}
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// 解法一 DFS + 二分
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func minimumEffortPath(heights [][]int) int {
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n, m := len(heights), len(heights[0])
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visited := make([][]bool, n)
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for i := range visited {
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visited[i] = make([]bool, m)
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}
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low, high := 0, 1000000
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for low < high {
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threshold := low + (high-low)>>1
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if !hasPath(heights, visited, 0, 0, threshold) {
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low = threshold + 1
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} else {
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high = threshold
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}
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for i := range visited {
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for j := range visited[i] {
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visited[i][j] = false
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}
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}
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}
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return low
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}
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func hasPath(heights [][]int, visited [][]bool, i, j, threshold int) bool {
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n, m := len(heights), len(heights[0])
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if i == n-1 && j == m-1 {
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return true
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}
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visited[i][j] = true
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res := false
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for _, d := range dir {
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ni, nj := i+d[0], j+d[1]
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if ni < 0 || ni >= n || nj < 0 || nj >= m || visited[ni][nj] || res {
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continue
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}
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diff := abs(heights[i][j] - heights[ni][nj])
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if diff <= threshold && hasPath(heights, visited, ni, nj, threshold) {
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res = true
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}
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}
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return res
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}
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func abs(a int) int {
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if a < 0 {
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a = -a
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}
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return a
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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func max(a, b int) int {
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if a < b {
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return b
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}
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return a
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}
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// 解法二 并查集
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func minimumEffortPath1(heights [][]int) int {
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n, m, edges, uf := len(heights), len(heights[0]), []edge{}, template.UnionFind{}
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uf.Init(n * m)
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for i, row := range heights {
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for j, h := range row {
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id := i*m + j
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if i > 0 {
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edges = append(edges, edge{id - m, id, abs(h - heights[i-1][j])})
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}
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if j > 0 {
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edges = append(edges, edge{id - 1, id, abs(h - heights[i][j-1])})
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}
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}
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}
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sort.Slice(edges, func(i, j int) bool { return edges[i].diff < edges[j].diff })
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for _, e := range edges {
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uf.Union(e.v, e.w)
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if uf.Find(0) == uf.Find(n*m-1) {
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return e.diff
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}
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}
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return 0
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}
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type edge struct {
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v, w, diff int
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}
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``` |