mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-06 09:23:19 +08:00
Add solution 417、916
This commit is contained in:
YDZ
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package leetcode
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import "math"
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func pacificAtlantic(matrix [][]int) [][]int {
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if len(matrix) == 0 || len(matrix[0]) == 0 {
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return nil
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}
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row, col, res := len(matrix), len(matrix[0]), make([][]int, 0)
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pacific, atlantic := make([][]bool, row), make([][]bool, row)
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for i := 0; i < row; i++ {
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pacific[i] = make([]bool, col)
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atlantic[i] = make([]bool, col)
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}
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for i := 0; i < row; i++ {
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dfs(matrix, i, 0, &pacific, math.MinInt32)
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dfs(matrix, i, col-1, &atlantic, math.MinInt32)
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}
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for j := 0; j < col; j++ {
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dfs(matrix, 0, j, &pacific, math.MinInt32)
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dfs(matrix, row-1, j, &atlantic, math.MinInt32)
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}
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for i := 0; i < row; i++ {
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for j := 0; j < col; j++ {
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if atlantic[i][j] && pacific[i][j] {
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res = append(res, []int{i, j})
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}
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}
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}
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return res
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}
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func dfs(matrix [][]int, row, col int, visited *[][]bool, height int) {
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if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) {
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return
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}
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if (*visited)[row][col] || matrix[row][col] < height {
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return
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}
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(*visited)[row][col] = true
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dfs(matrix, row+1, col, visited, matrix[row][col])
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dfs(matrix, row-1, col, visited, matrix[row][col])
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dfs(matrix, row, col+1, visited, matrix[row][col])
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dfs(matrix, row, col-1, visited, matrix[row][col])
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}
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@ -0,0 +1,56 @@
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question417 struct {
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para417
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ans417
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}
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// para 是参数
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// one 代表第一个参数
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type para417 struct {
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matrix [][]int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans417 struct {
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one [][]int
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}
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func Test_Problem417(t *testing.T) {
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qs := []question417{
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{
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para417{[][]int{
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{1, 2, 2, 3, 5},
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{3, 2, 3, 4, 4},
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{2, 4, 5, 3, 1},
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{6, 7, 1, 4, 5},
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{5, 1, 1, 2, 4},
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}},
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ans417{[][]int{
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{0, 4},
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{1, 3},
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{1, 4},
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{2, 2},
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{3, 0},
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{3, 1},
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{4, 0},
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}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 417------------------------\n")
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for _, q := range qs {
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_, p := q.ans417, q.para417
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fmt.Printf("【input】:%v 【output】:%v\n", p, pacificAtlantic(p.matrix))
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}
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fmt.Printf("\n\n\n")
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}
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92
leetcode/0417.Pacific-Atlantic-Water-Flow/README.md
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92
leetcode/0417.Pacific-Atlantic-Water-Flow/README.md
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# [417. Pacific Atlantic Water Flow](https://leetcode.com/problems/pacific-atlantic-water-flow/)
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## 题目
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Given an `m x n` matrix of non-negative integers representing the height of each unit cell in a continent, the "Pacific ocean" touches the left and top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges.
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Water can only flow in four directions (up, down, left, or right) from a cell to another one with height equal or lower.
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Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean.
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**Note:**
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1. The order of returned grid coordinates does not matter.
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2. Both m and n are less than 150.
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**Example:**
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```
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Given the following 5x5 matrix:
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Pacific ~ ~ ~ ~ ~
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~ 1 2 2 3 (5) *
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~ 3 2 3 (4) (4) *
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~ 2 4 (5) 3 1 *
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~ (6) (7) 1 4 5 *
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~ (5) 1 1 2 4 *
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* * * * * Atlantic
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Return:
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[[0, 4], [1, 3], [1, 4], [2, 2], [3, 0], [3, 1], [4, 0]] (positions with parentheses in above matrix).
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```
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## 题目大意
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给定一个 m x n 的非负整数矩阵来表示一片大陆上各个单元格的高度。“太平洋”处于大陆的左边界和上边界，而“大西洋”处于大陆的右边界和下边界。规定水流只能按照上、下、左、右四个方向流动，且只能从高到低或者在同等高度上流动。请找出那些水流既可以流动到“太平洋”，又能流动到“大西洋”的陆地单元的坐标。
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## 解题思路
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- 暴力解法，利用 DFS 把二维数据按照行优先搜索一遍，分别标记出太平洋和大西洋水流能到达的位置。再按照列优先搜索一遍，标记出太平洋和大西洋水流能到达的位置。最后两者都能到达的坐标即为所求。
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## 代码
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```go
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package leetcode
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import "math"
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func pacificAtlantic(matrix [][]int) [][]int {
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if len(matrix) == 0 || len(matrix[0]) == 0 {
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return nil
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}
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row, col, res := len(matrix), len(matrix[0]), make([][]int, 0)
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pacific, atlantic := make([][]bool, row), make([][]bool, row)
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for i := 0; i < row; i++ {
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pacific[i] = make([]bool, col)
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atlantic[i] = make([]bool, col)
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}
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for i := 0; i < row; i++ {
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dfs(matrix, i, 0, &pacific, math.MinInt32)
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dfs(matrix, i, col-1, &atlantic, math.MinInt32)
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}
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for j := 0; j < col; j++ {
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dfs(matrix, 0, j, &pacific, math.MinInt32)
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dfs(matrix, row-1, j, &atlantic, math.MinInt32)
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}
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for i := 0; i < row; i++ {
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for j := 0; j < col; j++ {
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if atlantic[i][j] && pacific[i][j] {
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res = append(res, []int{i, j})
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}
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}
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}
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return res
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}
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func dfs(matrix [][]int, row, col int, visited *[][]bool, height int) {
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if row < 0 || row >= len(matrix) || col < 0 || col >= len(matrix[0]) {
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return
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}
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if (*visited)[row][col] || matrix[row][col] < height {
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return
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}
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(*visited)[row][col] = true
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dfs(matrix, row+1, col, visited, matrix[row][col])
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dfs(matrix, row-1, col, visited, matrix[row][col])
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dfs(matrix, row, col+1, visited, matrix[row][col])
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dfs(matrix, row, col-1, visited, matrix[row][col])
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}
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```
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36
leetcode/0916.Word-Subsets/916. Word Subsets.go
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36
leetcode/0916.Word-Subsets/916. Word Subsets.go
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package leetcode
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func wordSubsets(A []string, B []string) []string {
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var counter [26]int
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for _, b := range B {
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var m [26]int
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for _, c := range b {
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j := c - 'a'
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m[j]++
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}
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for i := 0; i < 26; i++ {
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if m[i] > counter[i] {
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counter[i] = m[i]
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}
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}
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}
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var res []string
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for _, a := range A {
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var m [26]int
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for _, c := range a {
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j := c - 'a'
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m[j]++
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}
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ok := true
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for i := 0; i < 26; i++ {
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if m[i] < counter[i] {
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ok = false
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break
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}
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}
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if ok {
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res = append(res, a)
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}
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}
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return res
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}
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63
leetcode/0916.Word-Subsets/916. Word Subsets_test.go
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63
leetcode/0916.Word-Subsets/916. Word Subsets_test.go
Normal file
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question916 struct {
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para916
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ans916
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}
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// para 是参数
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// one 代表第一个参数
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type para916 struct {
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A []string
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B []string
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}
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// ans 是答案
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// one 代表第一个答案
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type ans916 struct {
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one []string
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}
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func Test_Problem916(t *testing.T) {
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qs := []question916{
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{
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para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"e", "o"}},
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ans916{[]string{"facebook", "google", "leetcode"}},
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},
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{
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para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"l", "e"}},
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ans916{[]string{"apple", "google", "leetcode"}},
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},
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{
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para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"e", "oo"}},
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ans916{[]string{"facebook", "google"}},
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},
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{
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para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"lo", "eo"}},
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ans916{[]string{"google", "leetcode"}},
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},
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{
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para916{[]string{"amazon", "apple", "facebook", "google", "leetcode"}, []string{"ec", "oc", "ceo"}},
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ans916{[]string{"facebook", "leetcode"}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 916------------------------\n")
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for _, q := range qs {
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_, p := q.ans916, q.para916
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fmt.Printf("【input】:%v 【output】:%v\n", p, wordSubsets(p.A, p.B))
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}
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fmt.Printf("\n\n\n")
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}
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103
leetcode/0916.Word-Subsets/README.md
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103
leetcode/0916.Word-Subsets/README.md
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# [916. Word Subsets](https://leetcode.com/problems/word-subsets/)
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## 题目
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We are given two arrays `A` and `B` of words. Each word is a string of lowercase letters.
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Now, say that word `b` is a subset of word `a` ****if every letter in `b` occurs in `a`, **including multiplicity**. For example, `"wrr"` is a subset of `"warrior"`, but is not a subset of `"world"`.
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Now say a word `a` from `A` is *universal* if for every `b` in `B`, `b` is a subset of `a`.
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Return a list of all universal words in `A`. You can return the words in any order.
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**Example 1:**
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```
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Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
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Output:["facebook","google","leetcode"]
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```
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**Example 2:**
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```
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Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
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Output:["apple","google","leetcode"]
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```
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**Example 3:**
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```
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Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
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Output:["facebook","google"]
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```
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**Example 4:**
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```
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Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
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Output:["google","leetcode"]
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```
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**Example 5:**
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```
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Input:A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
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Output:["facebook","leetcode"]
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```
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**Note:**
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1. `1 <= A.length, B.length <= 10000`
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2. `1 <= A[i].length, B[i].length <= 10`
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3. `A[i]` and `B[i]` consist only of lowercase letters.
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4. All words in `A[i]` are unique: there isn't `i != j` with `A[i] == A[j]`.
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## 题目大意
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我们给出两个单词数组 A 和 B。每个单词都是一串小写字母。现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称单词 b 是单词 a 的子集。 例如，“wrr” 是 “warrior” 的子集，但不是 “world” 的子集。如果对 B 中的每一个单词 b，b 都是 a 的子集，那么我们称 A 中的单词 a 是通用的。你可以按任意顺序以列表形式返回 A 中所有的通用单词。
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## 解题思路
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- 简单题。先统计出 B 数组中单词每个字母的频次，再在 A 数组中依次判断每个单词是否超过了这个频次，如果超过了即输出。
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## 代码
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```go
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package leetcode
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func wordSubsets(A []string, B []string) []string {
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var counter [26]int
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for _, b := range B {
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var m [26]int
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for _, c := range b {
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j := c - 'a'
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m[j]++
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}
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for i := 0; i < 26; i++ {
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if m[i] > counter[i] {
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counter[i] = m[i]
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}
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}
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}
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var res []string
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for _, a := range A {
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var m [26]int
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for _, c := range a {
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j := c - 'a'
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m[j]++
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}
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ok := true
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for i := 0; i < 26; i++ {
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if m[i] < counter[i] {
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ok = false
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break
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}
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}
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if ok {
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res = append(res, a)
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}
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}
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return res
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}
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```
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Reference in New Issue
Block a user