mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 08:27:30 +08:00
规范格式
This commit is contained in:
YDZ
149
leetcode/0542.01-Matrix/542. 01 Matrix.go
Normal file
149
leetcode/0542.01-Matrix/542. 01 Matrix.go
Normal file
@ -0,0 +1,149 @@
|
||||
package leetcode
|
||||
|
||||
import (
|
||||
"math"
|
||||
)
|
||||
|
||||
// 解法一 BFS
|
||||
func updateMatrixBFS(matrix [][]int) [][]int {
|
||||
res := make([][]int, len(matrix))
|
||||
if len(matrix) == 0 || len(matrix[0]) == 0 {
|
||||
return res
|
||||
}
|
||||
queue := make([][]int, 0)
|
||||
for i := range matrix {
|
||||
res[i] = make([]int, len(matrix[0]))
|
||||
for j := range res[i] {
|
||||
if matrix[i][j] == 0 {
|
||||
res[i][j] = -1
|
||||
queue = append(queue, []int{i, j})
|
||||
}
|
||||
}
|
||||
}
|
||||
level := 1
|
||||
for len(queue) > 0 {
|
||||
size := len(queue)
|
||||
for size > 0 {
|
||||
size--
|
||||
node := queue[0]
|
||||
queue = queue[1:]
|
||||
i, j := node[0], node[1]
|
||||
for _, direction := range [][]int{{-1, 0}, {1, 0}, {0, 1}, {0, -1}} {
|
||||
x := i + direction[0]
|
||||
y := j + direction[1]
|
||||
if x < 0 || x >= len(matrix) || y < 0 || y >= len(matrix[0]) || res[x][y] < 0 || res[x][y] > 0 {
|
||||
continue
|
||||
}
|
||||
res[x][y] = level
|
||||
queue = append(queue, []int{x, y})
|
||||
}
|
||||
}
|
||||
level++
|
||||
}
|
||||
for i, row := range res {
|
||||
for j, cell := range row {
|
||||
if cell == -1 {
|
||||
res[i][j] = 0
|
||||
}
|
||||
}
|
||||
}
|
||||
return res
|
||||
}
|
||||
|
||||
// 解法二 DFS
|
||||
func updateMatrixDFS(matrix [][]int) [][]int {
|
||||
result := [][]int{}
|
||||
if len(matrix) == 0 || len(matrix[0]) == 0 {
|
||||
return result
|
||||
}
|
||||
maxRow, maxCol := len(matrix), len(matrix[0])
|
||||
for r := 0; r < maxRow; r++ {
|
||||
for c := 0; c < maxCol; c++ {
|
||||
if matrix[r][c] == 1 && hasZero(matrix, r, c) == false {
|
||||
// 将四周没有 0 的 1 特殊处理为最大值
|
||||
matrix[r][c] = math.MaxInt64
|
||||
}
|
||||
}
|
||||
}
|
||||
for r := 0; r < maxRow; r++ {
|
||||
for c := 0; c < maxCol; c++ {
|
||||
if matrix[r][c] == 1 {
|
||||
dfsMatrix(matrix, r, c, -1)
|
||||
}
|
||||
}
|
||||
}
|
||||
return (matrix)
|
||||
}
|
||||
|
||||
// 判断四周是否有 0
|
||||
func hasZero(matrix [][]int, row, col int) bool {
|
||||
if row > 0 && matrix[row-1][col] == 0 {
|
||||
return true
|
||||
}
|
||||
if col > 0 && matrix[row][col-1] == 0 {
|
||||
return true
|
||||
}
|
||||
if row < len(matrix)-1 && matrix[row+1][col] == 0 {
|
||||
return true
|
||||
}
|
||||
if col < len(matrix[0])-1 && matrix[row][col+1] == 0 {
|
||||
return true
|
||||
}
|
||||
return false
|
||||
}
|
||||
|
||||
func dfsMatrix(matrix [][]int, row, col, val int) {
|
||||
// 不超过棋盘氛围，且 val 要比 matrix[row][col] 小
|
||||
if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) || (matrix[row][col] <= val) {
|
||||
return
|
||||
}
|
||||
if val > 0 {
|
||||
matrix[row][col] = val
|
||||
}
|
||||
dfsMatrix(matrix, row-1, col, matrix[row][col]+1)
|
||||
dfsMatrix(matrix, row, col-1, matrix[row][col]+1)
|
||||
dfsMatrix(matrix, row+1, col, matrix[row][col]+1)
|
||||
dfsMatrix(matrix, row, col+1, matrix[row][col]+1)
|
||||
}
|
||||
|
||||
// 解法三 DP
|
||||
func updateMatrixDP(matrix [][]int) [][]int {
|
||||
for i, row := range matrix {
|
||||
for j, val := range row {
|
||||
if val == 0 {
|
||||
continue
|
||||
}
|
||||
left, top := math.MaxInt16, math.MaxInt16
|
||||
if i > 0 {
|
||||
top = matrix[i-1][j] + 1
|
||||
}
|
||||
if j > 0 {
|
||||
left = matrix[i][j-1] + 1
|
||||
}
|
||||
matrix[i][j] = min(top, left)
|
||||
}
|
||||
}
|
||||
for i := len(matrix) - 1; i >= 0; i-- {
|
||||
for j := len(matrix[0]) - 1; j >= 0; j-- {
|
||||
if matrix[i][j] == 0 {
|
||||
continue
|
||||
}
|
||||
right, bottom := math.MaxInt16, math.MaxInt16
|
||||
if i < len(matrix)-1 {
|
||||
bottom = matrix[i+1][j] + 1
|
||||
}
|
||||
if j < len(matrix[0])-1 {
|
||||
right = matrix[i][j+1] + 1
|
||||
}
|
||||
matrix[i][j] = min(matrix[i][j], min(bottom, right))
|
||||
}
|
||||
}
|
||||
return matrix
|
||||
}
|
||||
|
||||
func min(a int, b int) int {
|
||||
if a > b {
|
||||
return b
|
||||
}
|
||||
return a
|
||||
}
|
||||
52
leetcode/0542.01-Matrix/542. 01 Matrix_test.go
Normal file
52
leetcode/0542.01-Matrix/542. 01 Matrix_test.go
Normal file
@ -0,0 +1,52 @@
|
||||
package leetcode
|
||||
|
||||
import (
|
||||
"fmt"
|
||||
"testing"
|
||||
)
|
||||
|
||||
type question542 struct {
|
||||
para542
|
||||
ans542
|
||||
}
|
||||
|
||||
// para 是参数
|
||||
// one 代表第一个参数
|
||||
type para542 struct {
|
||||
one [][]int
|
||||
}
|
||||
|
||||
// ans 是答案
|
||||
// one 代表第一个答案
|
||||
type ans542 struct {
|
||||
one [][]int
|
||||
}
|
||||
|
||||
func Test_Problem542(t *testing.T) {
|
||||
|
||||
qs := []question542{
|
||||
|
||||
question542{
|
||||
para542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{0, 0, 0}}},
|
||||
ans542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{0, 0, 0}}},
|
||||
},
|
||||
|
||||
question542{
|
||||
para542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{1, 1, 1}}},
|
||||
ans542{[][]int{[]int{0, 0, 0}, []int{0, 1, 0}, []int{1, 2, 1}}},
|
||||
},
|
||||
|
||||
question542{
|
||||
para542{[][]int{[]int{1, 1, 1, 1, 1, 1}, []int{1, 1, 1, 1, 1, 1}, []int{1, 1, 0, 0, 1, 1}, []int{1, 1, 0, 0, 1, 1}, []int{1, 1, 1, 1, 1, 1}, []int{1, 1, 1, 1, 1, 1}}},
|
||||
ans542{[][]int{[]int{4, 3, 2, 2, 3, 4}, []int{3, 2, 1, 1, 2, 3}, []int{2, 1, 0, 0, 1, 2}, []int{2, 1, 0, 0, 1, 2}, []int{3, 2, 1, 1, 2, 3}, []int{4, 3, 2, 2, 3, 4}}},
|
||||
},
|
||||
}
|
||||
|
||||
fmt.Printf("------------------------Leetcode Problem 542------------------------\n")
|
||||
|
||||
for _, q := range qs {
|
||||
_, p := q.ans542, q.para542
|
||||
fmt.Printf("【input】:%v 【output】:%v\n", p, updateMatrixDP(p.one))
|
||||
}
|
||||
fmt.Printf("\n\n\n")
|
||||
}
|
||||
52
leetcode/0542.01-Matrix/README.md
Executable file
52
leetcode/0542.01-Matrix/README.md
Executable file
@ -0,0 +1,52 @@
|
||||
# [542. 01 Matrix](https://leetcode.com/problems/01-matrix/)
|
||||
|
||||
|
||||
## 题目
|
||||
|
||||
Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
|
||||
|
||||
The distance between two adjacent cells is 1.
|
||||
|
||||
**Example 1:**
|
||||
|
||||
Input:
|
||||
[[0,0,0],
|
||||
[0,1,0],
|
||||
[0,0,0]]
|
||||
|
||||
Output:
|
||||
[[0,0,0],
|
||||
[0,1,0],
|
||||
[0,0,0]]
|
||||
|
||||
**Example 2:**
|
||||
|
||||
Input:
|
||||
[[0,0,0],
|
||||
[0,1,0],
|
||||
[1,1,1]]
|
||||
|
||||
Output:
|
||||
[[0,0,0],
|
||||
[0,1,0],
|
||||
[1,2,1]]
|
||||
|
||||
**Note:**
|
||||
|
||||
1. The number of elements of the given matrix will not exceed 10,000.
|
||||
2. There are at least one 0 in the given matrix.
|
||||
3. The cells are adjacent in only four directions: up, down, left and right.
|
||||
|
||||
|
||||
## 题目大意
|
||||
|
||||
给定一个由 0 和 1 组成的矩阵，找出每个元素到最近的 0 的距离。两个相邻元素间的距离为 1 。
|
||||
|
||||
|
||||
## 解题思路
|
||||
|
||||
|
||||
- 给出一个二维数组，数组里面只有 0 和 1 。要求计算每个 1 距离最近的 0 的距离。
|
||||
- 这一题有 3 种解法，第一种解法最容易想到，BFS。先预处理一下棋盘，将每个 0 都处理为 -1 。将 1 都处理为 0 。将每个 -1 (即原棋盘的 0)都入队，每次出队都将四周的 4 个位置都入队。这就想一颗石头扔进了湖里，一圈一圈的波纹荡开，每一圈都是一层。由于棋盘被我们初始化了，所有为 -1 的都是原来为 0 的，所以波纹扫过来不需要处理这些 -1 的点。棋盘上为 0 的点都是原来为 1 的点，这些点在波纹扫过来的时候就需要赋值更新 level。当下次波纹再次扫到原来为 1 的点的时候，由于它已经被第一次到的波纹更新了值，所以这次不用再更新了。(第一次波纹到的时候一定是最短的)
|
||||
- 第二种解法是 DFS。先预处理，把周围没有 0 的 1 都重置为最大值。当周围有 0 的 1，距离 0 的位置都是 1，这些点是不需要动的，需要更新的点恰恰应该是那些周围没有 0 的点。当递归的步数 val 比点的值小(这也就是为什么会先把 1 更新成最大值的原因)的时候，不断更新它。
|
||||
- 第三种解法是 DP。由于有 4 个方向，每次处理 2 个方向，可以降低时间复杂度。第一次循环从上到下，从左到右遍历，先处理上边和左边，第二次循环从下到上，从右到左遍历，再处理右边和下边。
|
||||
Reference in New Issue
Block a user