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
}