算法61天|图论
leetcode-master/图论广索理论基础.md at master · youngyangyang04/leetcode-master · GitHub
leetcode-master/图论深搜理论基础.md at master · youngyangyang04/leetcode-master · GitHub
力扣
1791. 找出星型图的中心节点
var findCenter = function(edges) {
const n = edges.length + 1;
const degree = new Array(n+1).fill(0)
for(const edge of edges){
degree[edge[0]]++
degree[edge[1]]++
}
for(let i = 0;;i++){
if(degree[i] === n-1){
return i
}
}
};统计各个节点的度，使用数组的方法统计，将节点的值当作数组的索引。然后有多一个的话，就++。会发现中间节点的度是n-1，其它的节点的度是1
tleetcode-master/1791.找出星型图的中心节点.md at master · youngyangyang04/leetcode-master · GitHub
797. 所有可能的路径
var allPathsSourceTarget = function(graph) {
let result = []
let path = []
//x 是当前遍历的节点
// graph 是存当前的图
const dfs = function(path,graph,x){
// 终止条件
if(x === graph.length - 1){
//将结果加入
result.push([...path])
return
}
for(let i = 0;i力扣
200. 岛屿数量
dfs
var numIslands = function (grid) {
// 使用dfs
let dirx = [-1, 1, 0, 0]
let diry = [0, 0, -1, 1]
let row = grid.length
let col = grid[0].length
const dfs = function (grid, visited, i, j) {
//终止条件
//for循环
for (let k = 0; k < 4; k++) {
let x = i + dirx[k]
let y = j + diry[k]
if (x >= row || y >= col || x < 0 || y < 0) {
continue
}
if (visited[x][y] === false && grid[x][y] === '1') {
visited[x][y] = true
dfs(grid, visited, x, y)
}
}
}
let visited = new Array(grid.length).fill(false).map(() => new Array(grid[0].length).fill(false))
let count = 0
for (let i = 0; i < row; i++) {
for (let j = 0; j < col; j++) {
if (visited[i][j] === false && grid[i][j] === '1') {
visited[i][j] === true
count++;
dfs(grid, visited, i, j)
}
}
}
return count
};BFS
var numIslands = function (grid) {
let dirx = [-1, 1, 0, 0]
let diry = [0, 0, -1, 1]
let row = grid.length
let col = grid[0].length
let visited = new Array(row).fill(false).map(() => new Array(col).fill(false))
// 先将节点加入队列中
const bfs = function (grid,visited,i,j) {
let que = []
que.push([i, j])
visited[i][j] = true
while (que.length > 0) {
// 弹出当前节点,获取当前元素的坐标
const [curx, cury] = que.shift()
//判断它上下左右有没有岛屿
for (let k = 0; k < 4; k++) {
let x = curx + dirx[k]
let y = cury + diry[k]
if (x < 0 || y < 0 || x >= row || y >= col) continue
if (visited[x][y] != true && grid[x][y] === '1') {
visited[x][y] = true
//加入节点
que.push([x, y])
}
}
}
}
let count = 0
for(let i = 0;i
463. 岛屿的周长
如果碰到水域的话就+1，如果碰到边界也加1
var islandPerimeter = function(grid) {
let row = grid.length
let col = grid[0].length
// i-1,j/i+1,j/i,j-1/i,j+1
let dirx = [-1,1,0,0]
let diry = [0,0,-1,1]
let count = 0
for(let i = 0;i= row ||
y >= col||
x < 0||
y<0||
grid[x][y] === 0){
count ++;
}
}
// 判断上下左右，如果遇到水域+1，如果遇到边界就+1
}
}
}
return count
};
695. 岛屿的最大面积
BFS
var maxAreaOfIsland = function (grid) {
const bfs = function (grid, i, j, visited,count) {
const dirx = [-1, 1, 0, 0]
const diry = [0, 0, -1, 1]
let que = []
visited[i][j] = true
que.push([i, j])
count ++
while (que.length > 0) {
const [curx, cury] = que.shift()
for (let k = 0; k < 4; k++) {
let newx = curx + dirx[k]
let newy = cury + diry[k]
if (newx < 0 || newy < 0|| newx >= row || newy >= col) continue
if (grid[newx][newy] === 1 && visited[newx][newy] !== true) {
visited[newx][newy] = true
que.push([newx,newy])
count ++
}
}
}
return count
}
// 找到岛屿，找到的话，就数统计个数
let result = 0
// bfs
const col = grid[0].length
const row = grid.length
let visited = new Array(row).fill(false).map(() => new Array(col).fill(false))
for (let i = 0; i < row; i++) {
for (let j = 0; j < col; j++) {
if (visited[i][j] === false && grid[i][j] === 1) {
let count = 0
count = bfs(grid, i, j, visited,count)
result = Math.max(result,count)
}
}
}
return result
};dfs
var maxAreaOfIsland = function(grid) {
// This function uses depth-first search algorithm to traverse the binary matrix grid.
const dfs = function(grid,i,j,visited){
const dirx = [-1,1,0,0]
const diry = [0,0,-1,1]
visited[i][j] = true
let area = 1
for(let k= 0;k<4;k++){
let curx = dirx[k] + i
let cury = diry[k] + j
if(curx < 0 || cury <0 || curx >= row || cury >=col) continue
if(visited[curx][cury] === false && grid[curx][cury] === 1){
visited[curx][cury] = true
area += dfs(grid,curx,cury,visited)
}
}
return area
}
const row = grid.length
const col = grid[0].length
let result = 0
let visited = new Array(row).fill(false).map(()=>new Array(col).fill(false))
for(let i = 0;i
827. 最大人工岛(hard)
1020. 飞地的数量 (medium)
var numEnclaves = function(grid) {
//bfs
const bfs = function(grid,i,j,visited,count){
let que = []
const dirx = [-1,1,0,0]
const diry = [0,0,-1,1]
que.push([i,j])
count ++
visited[i][j] = true
while(que.length > 0){
const [x,y] = que.shift()
for(let k = 0;k<4;k++){
let curx = x + dirx[k]
let cury = y + diry[k]
if(curx < 0 || cury < 0|| curx>=row || cury >= col) continue
if(visited[curx][cury] === false && grid[curx][cury] === 1){
que.push([curx,cury])
count ++
visited[curx][cury] = true
}
}
}
return count
}
const row = grid.length
const col = grid[0].length
let result = 0
let visited = new Array(row).fill(false).map(()=>new Array(col).fill(false))
for(let i = 0;i这道题是先把边界的东西全部转化为true，但是并不需要给他计数。
130. 被围绕的区域
var solve = function(board) {
const bfs = function(board,i,j){
const dirx = [-1,1,0,0]
const diry = [0,0,-1,1]
let que = []
que.push([i,j])
board[i][j] = "A"
while(que.length>0){
let [x,y] = que.shift()
for(let k = 0;k<4;k++){
let curx = dirx[k]+x
let cury = diry[k]+y
if(curx<0||cury<0||curx>=row||cury>=col) {continue}
if( board[curx][cury] === "O"){
que.push([curx,cury])
board[curx][cury] = "A"
}
}
}
}
const row = board.length
const col = board[0].length
let visited = new Array(row).fill(false).map(()=>new Array(col).fill(false))
// 把边缘的都从O转化为了A
for(let i = 0;i这道题和飞地数量那道题一样的思路。需要注意的地方是，我们需要多次访问数组中的值，所以我们不能使用visited来标记，不然的话没有办法再遍历一次。
需要先把边缘的岛屿转化为A，然后第二部再遍历一次，如果遇到A就转化为O,如果遇到O就转化为A。
这些题都好容易写错，写=== 的时候很容易写成=
var solve = function (board) {
const dfs = function (board, i, j) {
const dirx = [-1, 1, 0, 0]
const diry = [0, 0, -1, 1]
board[i][j] = "A"
for (let k = 0; k < 4; k++) {
let curx = dirx[k] + i
let cury = diry[k] + j
if (curx < 0 || cury < 0 || curx >= row || cury >= col) { continue }
if (board[curx][cury] === "O") {
board[curx][cury] = "A"
dfs(board, curx, cury)
}
}
}
const row = board.length
const col = board[0].length
let visited = new Array(row).fill(false).map(() => new Array(col).fill(false))
// 把边缘的都从O转化为了A
for (let i = 0; i < row; i++) {
if (board[i][0] === "O") dfs(board, i, 0)
if (board[i][col - 1] === "O") dfs(board, i, col - 1)
}
console.log(board)
for (let i = 0; i < col; i++) {
if (board[0][i] === "O") {
dfs(board, 0, i)
}
if (board[row - 1][i] === "O") {
dfs(board, row - 1, i)
}
}
console.log(board)
for (let i = 0; i < row; i++) {
for (let j = 0; j < col; j++) {
if (board[i][j] === "O") {
board[i][j] = "X"
}
if (board[i][j] === "A") {
board[i][j] = "O"
}
}
}
return board
};
417. 太平洋大西洋水流问题(深搜、广搜)
var numIslands = function (grid) {
// 使用dfs
let dirx = [-1, 1, 0, 0]
let diry = [0, 0, -1, 1]
let row = grid.length
let col = grid[0].length
const dfs = function (grid, visited, i, j) {
//终止条件
//for循环
for (let k = 0; k < 4; k++) {
let x = i + dirx[k]
let y = j + diry[k]
if (x >= row || y >= col || x < 0 || y < 0) {
continue
}
if (visited[x][y] === false && grid[x][y] === '1') {
visited[x][y] = true
dfs(grid, visited, x, y)
}
}
}
let visited = new Array(grid.length).fill(false).map(() => new Array(grid[0].length).fill(false))
let count = 0
for (let i = 0; i < row; i++) {
for (let j = 0; j < col; j++) {
if (visited[i][j] === false && grid[i][j] === '1') {
visited[i][j] === true
count++;
dfs(grid, visited, i, j)
}
}
}
return count
};
127. 单词接龙 （困难）
841. 钥匙和房间
bfs
var canVisitAllRooms = function(rooms) {
// 使用dfs深度优先算法,使用visited记录访问过的屋子,使用key拿到钥匙
const dfs = function(visited,room,key){
// 终止条件
if(visited[key]){
return
}
// 单层递归逻辑
visited[key] = true
let keys = room[key]
for(let i = 0;idfs
var canVisitAllRooms = function(rooms) {
// 使用dfs深度优先算法,使用visited记录访问过的屋子,使用key拿到钥匙
const bfs = function(visited,room,key){
// 终止条件
let que = []
que.push(room[key])
// 单层递归逻辑
visited[key] = true
while(que.length>0){
let keys = que.shift()
for(let i = 0;i
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