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) = 3.
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 <= 10^4
s[i] and c are lowercase English letters.
It is guaranteed that c occurs at least once in s.

Solution

Time complexity : O(n^2)
Space complexity : O(n)

class Solution {
public:
    vector<int> shortestToChar(string s, char c) {
        int n = s.size();
        vector<int> ans(n, INT_MAX);
        for (int i=0; i<n; ++i) {
            if (s[i] != c) continue;
            ans[i] = 0;
            for (int j=0; j<n; ++j)
                ans[j] = min(ans[j], abs(j-i));
        }
        return ans;
    }
};

用一 vector 記錄最短距離，遍歷元素，每遇到 c 時，就更新最小距離值。