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feat: 关于图第1971题js代码
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@ -14,20 +14,18 @@
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给你数组 edges 和整数 n、start 和 end,如果从 start 到 end 存在 有效路径 ,则返回 true,否则返回 false 。
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提示:
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* 1 <= n <= 2 * 10^5
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* 0 <= edges.length <= 2 * 10^5
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* edges[i].length == 2
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* 0 <= ui, vi <= n - 1
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* ui != vi
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* 0 <= start, end <= n - 1
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* 不存在双向边
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* 不存在指向顶点自身的边
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- 1 <= n <= 2 \* 10^5
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- 0 <= edges.length <= 2 \* 10^5
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- edges[i].length == 2
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- 0 <= ui, vi <= n - 1
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- ui != vi
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- 0 <= start, end <= n - 1
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- 不存在双向边
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- 不存在指向顶点自身的边
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## 思路
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@ -70,7 +68,7 @@ void join(int u, int v) {
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}
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```
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以上模板中,只要修改 n 大小就可以,本题n不会超过2 * 10^5。
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以上模板中,只要修改 n 大小就可以,本题 n 不会超过 2 \* 10^5。
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并查集主要有三个功能。
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@ -206,6 +204,85 @@ class Solution:
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return find(source) == find(destination)
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```
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### Javascript:
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Javascript 并查集解法如下:
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```Javascript
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class unionF{
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constructor(n){
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this.count = n
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this.roots = new Array(n).fill(0).map((item,index)=>index)
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}
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findRoot(x){
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if(this.roots[x]!==x){
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this.roots[x] = this.findRoot(this.roots[x])
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}
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return this.roots[x]
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}
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union(x,y){
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const rx = this.findRoot(x)
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const ry = this.findRoot(y)
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this.roots[rx] = ry
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this.count--
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}
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isConnected(x,y){
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return this.findRoot(x)===this.findRoot(y)
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}
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}
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var validPath = function(n, edges, source, destination) {
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const UF = new unionF(n)
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for(const [s,t] of edges){
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UF.union(s,t)
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}
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return UF.isConnected(source,destination)
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};
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```
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Javascript 双向 bfs 解法如下:
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```Javascript
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var validPath = function(n, edges, source, destination) {
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const graph = new Array(n).fill(0).map(()=>[])
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for(const [s,t] of edges){
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graph[s].push(t)
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graph[t].push(s)
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}
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const visited = new Array(n).fill(false)
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function bfs(start,end,graph){
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const startq = [start]
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const endq = [end]
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while(startq.length&&endq.length){
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const slen = startq.length
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for(let i = 0;i<slen;i++){
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const scur = startq.shift()
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if(visited[scur]) continue
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if(endq.includes(scur)) return true
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visited[scur] = true
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const neighbors = graph[scur]
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startq.push(...neighbors)
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}
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const elen = endq.length
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for(let i = 0;i<elen;i++){
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const ecur = endq.shift()
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if(visited[ecur]) continue
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if(startq.includes(ecur)) return true
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visited[ecur] = true
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const neighbors = graph[ecur]
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endq.push(...neighbors)
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}
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}
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return false
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}
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return bfs(source,destination,graph)
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};
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```
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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