public class Solution
extends Object
821 - Shortest Distance to a Character\.
Easy
Given a string `s` and a character `c` that occurs in `s`, return _an array of integers_ `answer` _where_ `answer.length == s.length` _and_ `answer[i]` _is the **distance** from index_ `i` _to the **closest** occurrence of character_ `c` _in_ `s`.
The **distance** between two indices `i` and `j` is `abs(i - j)`, where `abs` is the absolute value function.
**Example 1:**
**Input:** s = "loveleetcode", c = "e"
**Output:** [3,2,1,0,1,0,0,1,2,2,1,0]
**Explanation:** The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed).
The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2.
For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.
**Example 2:**
**Input:** s = "aaab", c = "b"
**Output:** [3,2,1,0]
**Constraints:**
* 1 <= s.length <= 104
* `s[i]` and `c` are lowercase English letters.
* It is guaranteed that `c` occurs at least once in `s`.