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feat: add typescript solution to lc problem: No.0821

No.0821.Shortest Distance to a Character

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Original file line number	Diff line number	Diff line change
`@@ -65,6 +65,26 @@`
`65`	`65`
`66`	`66`	```
`67`	`67`
	`68`	`+### TypeScript`
	`69`	`+`
	`70`	+```ts
	`71`	`+function shortestToChar(s: string, c: string): number[] {`
	`72`	`+ const n = s.length;`
	`73`	`+ let ans = [];`
	`74`	`+ let pre = Infinity;`
	`75`	`+ for (let i = 0; i < n; i++) {`
	`76`	`+ if (s.charAt(i) == c) pre = i;`
	`77`	`+ ans[i] = Math.abs(pre - i);`
	`78`	`+ }`
	`79`	`+ pre = Infinity;`
	`80`	`+ for (let i = n - 1; i > -1; i--) {`
	`81`	`+ if (s.charAt(i) == c) pre = i;`
	`82`	`+ ans[i] = Math.min(Math.abs(pre - i), ans[i]);`
	`83`	`+ }`
	`84`	`+ return ans;`
	`85`	`+};`
	`86`	+```
	`87`	`+`
`68`	`88`	`### ...`
`69`	`89`
`70`	`90`	```

Lines changed: 20 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -54,6 +54,26 @@ The closest occurrence of 'e' for index 8 is at index 6, so the distance`
`54`	`54`
`55`	`55`	```
`56`	`56`
	`57`	`+### TypeScript`
	`58`	`+`
	`59`	+```ts
	`60`	`+function shortestToChar(s: string, c: string): number[] {`
	`61`	`+ const n = s.length;`
	`62`	`+ let ans = [];`
	`63`	`+ let pre = Infinity;`
	`64`	`+ for (let i = 0; i < n; i++) {`
	`65`	`+ if (s.charAt(i) == c) pre = i;`
	`66`	`+ ans[i] = Math.abs(pre - i);`
	`67`	`+ }`
	`68`	`+ pre = Infinity;`
	`69`	`+ for (let i = n - 1; i > -1; i--) {`
	`70`	`+ if (s.charAt(i) == c) pre = i;`
	`71`	`+ ans[i] = Math.min(Math.abs(pre - i), ans[i]);`
	`72`	`+ }`
	`73`	`+ return ans;`
	`74`	`+};`
	`75`	+```
	`76`	`+`
`57`	`77`	`### ...`
`58`	`78`
`59`	`79`	```

Lines changed: 15 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,15 @@`
	`1`	`+function shortestToChar(s: string, c: string): number[] {`
	`2`	`+ const n = s.length;`
	`3`	`+ let ans = [];`
	`4`	`+ let pre = Infinity;`
	`5`	`+ for (let i = 0; i < n; i++) {`
	`6`	`+ if (s.charAt(i) == c) pre = i;`
	`7`	`+ ans[i] = Math.abs(pre - i);`
	`8`	`+ }`
	`9`	`+ pre = Infinity;`
	`10`	`+ for (let i = n - 1; i > -1; i--) {`
	`11`	`+ if (s.charAt(i) == c) pre = i;`
	`12`	`+ ans[i] = Math.min(Math.abs(pre - i), ans[i]);`
	`13`	`+ }`
	`14`	`+ return ans;`
	`15`	`+};`

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