# 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`.

### 解題

Runtime: 0 ms, faster than 100.00%
Memory Usage: 2.4 MB, less than 36.59%
func shortestToChar(s string, c byte) []int {
length := len(s)
ans := make([]int, length)
for i:=0; i<len(ans); i++ {
if s[i]==c {
for j:=i+1; j<len(ans); j++ {
if s[j]== c { break }
if ans[j]==0 { ans[j] = j-i }
ans[j] = min(j-i, ans[j])
}
for j:=i-1; j>=0; j-- {
if s[j]== c { break }
if ans[j]==0 { ans[j] = i-j }
ans[j] = min(i-j, ans[j])
}
}
}
return ans
}
func min(a, b int) int {
if a<b { return a }
return b
}