LeetCode-in-Java
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 <= 104s[i]andcare lowercase English letters.- It is guaranteed that
coccurs at least once ins.
Solution
import java.util.Arrays;
public class Solution {
public int[] shortestToChar(String s, char c) {
int[] result = new int[s.length()];
Arrays.fill(result, Integer.MAX_VALUE);
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == c) {
result[i] = 0;
}
}
for (int i = 0; i < s.length(); i++) {
if (result[i] != 0) {
int j = i - 1;
while (j >= 0 && result[j] != 0) {
j--;
}
if (j >= 0) {
result[i] = i - j;
}
j = i + 1;
while (j < s.length() && result[j] != 0) {
j++;
}
if (j < s.length()) {
result[i] = Math.min(result[i], j - i);
}
}
}
return result;
}
}