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feat: add typescript solution to lc problem: No.2065 (doocs#613)

No.2065.Maximum Path Quality of a Graph

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Original file line number	Diff line number	Diff line change
`@@ -84,6 +84,8 @@`
`84`	`84`
`85`	`85`	`<!-- 这里可写通用的实现逻辑 -->`
`86`	`86`
	`87`	`+DFS`
	`88`	`+`
`87`	`89`	`<!-- tabs:start -->`
`88`	`90`
`89`	`91`	`### Python3`
`@@ -102,6 +104,46 @@`
`102`	`104`
`103`	`105`	```
`104`	`106`
	`107`	`+### TypeScript`
	`108`	`+`
	`109`	+```ts
	`110`	`+function maximalPathQuality(values: number[], edges: number[][], maxTime: number): number {`
	`111`	`+ const n = values.length;`
	`112`	`+ let g: Array<Array<Array<number>>> = Array.from({ length: n }, v => new Array());`
	`113`	`+ for (let edge of edges) {`
	`114`	`+ let [u, v, t] = edge;`
	`115`	`+ g[u].push([v, t]);`
	`116`	`+ g[v].push([u, t]);`
	`117`	`+ }`
	`118`	`+ let visited = new Array(n).fill(false);`
	`119`	`+ let ans = 0;`
	`120`	`+`
	`121`	`+ function dfs(u: number, time: number, value: number): void {`
	`122`	`+ // 索引0为终点`
	`123`	`+ if (!u) {`
	`124`	`+ ans = Math.max(value, ans);`
	`125`	`+ }`
	`126`	`+ for (let [v, dist] of g[u]) {`
	`127`	`+ if (time - dist >= 0) {`
	`128`	`+ if (!visited[v]) {`
	`129`	`+ visited[v] = true;`
	`130`	`+ dfs(v, time - dist, value + values[v]);`
	`131`	`+ visited[v] = false; // 回溯`
	`132`	`+ } else {`
	`133`	`+ dfs(v, time - dist, value);`
	`134`	`+ }`
	`135`	`+ }`
	`136`	`+ }`
	`137`	`+ }`
	`138`	`+`
	`139`	`+ // 索引0为起点`
	`140`	`+ visited[0] = true;`
	`141`	`+ dfs(0, maxTime, values[0]);`
	`142`	`+`
	`143`	`+ return ans;`
	`144`	`+};`
	`145`	+```
	`146`	`+`
`105`	`147`	`### ...`
`106`	`148`
`107`	`149`	```

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Original file line number	Diff line number	Diff line change
`@@ -86,6 +86,46 @@ The only node visited is 0, giving a maximal path quality of 0.`
`86`	`86`
`87`	`87`	```
`88`	`88`
	`89`	`+### TypeScript`
	`90`	`+`
	`91`	+```ts
	`92`	`+function maximalPathQuality(values: number[], edges: number[][], maxTime: number): number {`
	`93`	`+ const n = values.length;`
	`94`	`+ let g: Array<Array<Array<number>>> = Array.from({ length: n }, v => new Array());`
	`95`	`+ for (let edge of edges) {`
	`96`	`+ let [u, v, t] = edge;`
	`97`	`+ g[u].push([v, t]);`
	`98`	`+ g[v].push([u, t]);`
	`99`	`+ }`
	`100`	`+ let visited = new Array(n).fill(false);`
	`101`	`+ let ans = 0;`
	`102`	`+`
	`103`	`+ function dfs(u: number, time: number, value: number): void {`
	`104`	`+ // 索引0为终点`
	`105`	`+ if (!u) {`
	`106`	`+ ans = Math.max(value, ans);`
	`107`	`+ }`
	`108`	`+ for (let [v, dist] of g[u]) {`
	`109`	`+ if (time - dist >= 0) {`
	`110`	`+ if (!visited[v]) {`
	`111`	`+ visited[v] = true;`
	`112`	`+ dfs(v, time - dist, value + values[v]);`
	`113`	`+ visited[v] = false; // 回溯`
	`114`	`+ } else {`
	`115`	`+ dfs(v, time - dist, value);`
	`116`	`+ }`
	`117`	`+ }`
	`118`	`+ }`
	`119`	`+ }`
	`120`	`+`
	`121`	`+ // 索引0为起点`
	`122`	`+ visited[0] = true;`
	`123`	`+ dfs(0, maxTime, values[0]);`
	`124`	`+`
	`125`	`+ return ans;`
	`126`	`+};`
	`127`	+```
	`128`	`+`
`89`	`129`	`### ...`
`90`	`130`
`91`	`131`	```

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Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,35 @@`
	`1`	`+function maximalPathQuality(values: number[], edges: number[][], maxTime: number): number {`
	`2`	`+ const n = values.length;`
	`3`	`+ let g: Array<Array<Array<number>>> = Array.from({ length: n }, v => new Array());`
	`4`	`+ for (let edge of edges) {`
	`5`	`+ let [u, v, t] = edge;`
	`6`	`+ g[u].push([v, t]);`
	`7`	`+ g[v].push([u, t]);`
	`8`	`+ }`
	`9`	`+ let visited = new Array(n).fill(false);`
	`10`	`+ let ans = 0;`
	`11`	`+`
	`12`	`+ function dfs(u: number, time: number, value: number): void {`
	`13`	`+ // 索引0为终点`
	`14`	`+ if (!u) {`
	`15`	`+ ans = Math.max(value, ans);`
	`16`	`+ }`
	`17`	`+ for (let [v, dist] of g[u]) {`
	`18`	`+ if (time - dist >= 0) {`
	`19`	`+ if (!visited[v]) {`
	`20`	`+ visited[v] = true;`
	`21`	`+ dfs(v, time - dist, value + values[v]);`
	`22`	`+ visited[v] = false; // 回溯`
	`23`	`+ } else {`
	`24`	`+ dfs(v, time - dist, value);`
	`25`	`+ }`
	`26`	`+ }`
	`27`	`+ }`
	`28`	`+ }`
	`29`	`+`
	`30`	`+ // 索引0为起点`
	`31`	`+ visited[0] = true;`
	`32`	`+ dfs(0, maxTime, values[0]);`
	`33`	`+`
	`34`	`+ return ans;`
	`35`	`+};`

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