From 200b30895ce8a0918c9e9c161c734cad3e4a625f Mon Sep 17 00:00:00 2001 From: halfrost Date: Thu, 1 Sep 2022 20:20:20 +0800 Subject: [PATCH] Update 1293 solution --- ...nary-Number-in-a-Linked-List-to-Integer.md | 2 +- ...th-in-a-Grid-with-Obstacles-Elimination.md | 138 ++++++++++++++++++ ...Find-Numbers-with-Even-Number-of-Digits.md | 2 +- 3 files changed, 140 insertions(+), 2 deletions(-) create mode 100644 website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md diff --git a/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md b/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md index 41633eae..af3a2ca0 100644 --- a/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md +++ b/website/content/ChapterFour/1200~1299/1290.Convert-Binary-Number-in-a-Linked-List-to-Integer.md @@ -84,5 +84,5 @@ func getDecimalValue(head *ListNode) int { ----------------------------------------------

⬅️上一页

-

下一页➡️

+

下一页➡️

diff --git a/website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md b/website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md new file mode 100644 index 00000000..f4cebe1f --- /dev/null +++ b/website/content/ChapterFour/1200~1299/1293.Shortest-Path-in-a-Grid-with-Obstacles-Elimination.md @@ -0,0 +1,138 @@ +# [1293. Shortest Path in a Grid with Obstacles Elimination](https://leetcode.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/) + + + +## 题目 + +You are given an m x n integer matrix grid where each cell is either 0 (empty) or 1 (obstacle). You can move up, down, left, or right from and to an empty cell in one step. + +Return the minimum number of steps to walk from the upper left corner (0, 0) to the lower right corner (m - 1, n - 1) given that you can eliminate at most k obstacles. If it is not possible to find such walk return -1. + + + +Example 1: + + +![](https://assets.leetcode.com/uploads/2021/09/30/short1-grid.jpg) + + +``` +Input: grid = [[0,0,0],[1,1,0],[0,0,0],[0,1,1],[0,0,0]], k = 1 +Output: 6 +Explanation: +The shortest path without eliminating any obstacle is 10. +The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2). +``` + +Example 2: + +![](https://assets.leetcode.com/uploads/2021/09/30/short2-grid.jpg) + +``` +Input: grid = [[0,1,1],[1,1,1],[1,0,0]], k = 1 +Output: -1 +Explanation: We need to eliminate at least two obstacles to find such a walk. +``` + +Constraints: + +- m == grid.length +- n == grid[i].length +- 1 <= m, n <= 40 +- 1 <= k <= m * n +- grid[i][j] is either 0 or 1. +- grid[0][0] == grid[m - 1][n - 1] == 0 + + + +## 题目大意 + +给你一个 m * n 的网格，其中每个单元格不是 0（空）就是 1（障碍物）。每一步，您都可以在空白单元格中上、下、左、右移动。 + +如果您 最多 可以消除 k 个障碍物，请找出从左上角 (0, 0) 到右下角 (m-1, n-1) 的最短路径，并返回通过该路径所需的步数。如果找不到这样的路径，则返回 -1 。 + + +## 解题思路 + +使用 BFS 遍历棋盘。这题比普通可达性问题多了一个障碍物的限制。这个也不难。每个点往周边四个方向扩展的时候，如果遇到障碍物，先算上这个障碍物，障碍物累积总个数小于 K 的时候，从障碍物的这个格子继续开始遍历。如果没有遇到障碍物，判断当前累积障碍物个数是否已经小于 K 个，如果小于 K 便继续遍历。如果大于 K，便终止此轮遍历。 + +## 代码 + +```go +var dir = [][]int{ + {-1, 0}, + {0, 1}, + {1, 0}, + {0, -1}, +} + +type pos struct { + x, y int + obstacle int + step int +} + +func shortestPath(grid [][]int, k int) int { + queue, m, n := []pos{}, len(grid), len(grid[0]) + visitor := make([][][]int, m) + if len(grid) == 1 && len(grid[0]) == 1 { + return 0 + } + for i := 0; i < m; i++ { + visitor[i] = make([][]int, n) + for j := 0; j < n; j++ { + visitor[i][j] = make([]int, k+1) + } + } + visitor[0][0][0] = 1 + queue = append(queue, pos{x: 0, y: 0, obstacle: 0, step: 0}) + for len(queue) > 0 { + size := len(queue) + for size > 0 { + size-- + node := queue[0] + queue = queue[1:] + for i := 0; i < len(dir); i++ { + newX := node.x + dir[i][0] + newY := node.y + dir[i][1] + if newX == m-1 && newY == n-1 { + if node.obstacle != 0 { + if node.obstacle <= k { + return node.step + 1 + } else { + continue + } + } + return node.step + 1 + } + if isInBoard(grid, newX, newY) { + if grid[newX][newY] == 1 { + if node.obstacle+1 <= k && visitor[newX][newY][node.obstacle+1] != 1 { + queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle + 1, step: node.step + 1}) + visitor[newX][newY][node.obstacle+1] = 1 + } + } else { + if node.obstacle <= k && visitor[newX][newY][node.obstacle] != 1 { + queue = append(queue, pos{x: newX, y: newY, obstacle: node.obstacle, step: node.step + 1}) + visitor[newX][newY][node.obstacle] = 1 + } + } + + } + } + } + } + return -1 +} + +func isInBoard(board [][]int, x, y int) bool { + return x >= 0 && x < len(board) && y >= 0 && y < len(board[0]) +} +``` + + +---------------------------------------------- +
+

⬅️上一页

+

下一页➡️

+
diff --git a/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md b/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md index f9e0d8f3..afb61361 100644 --- a/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md +++ b/website/content/ChapterFour/1200~1299/1295.Find-Numbers-with-Even-Number-of-Digits.md @@ -65,6 +65,6 @@ func findNumbers(nums []int) int { ----------------------------------------------
-

⬅️上一页

+

⬅️上一页

下一页➡️