Shortest Distance to a Character
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.
思路一：遍历两遍，先向前遍历，然后向后遍历，计算最小距离
public int[] shortestToChar(String s, char c) {
int[] ans = new int[s.length()];
char[] chars = s.toCharArray();
int gap = 0;
for (int i = 0; i < chars.length; i++) {
if (chars[i] == c) {
ans[i] = 0;
gap = 1;
} else {
if (gap > 0) {
ans[i] = gap++;
}
}
}
gap = 0;
for (int i = chars.length - 1; i >= 0; i--) {
if (chars[i] == c) {
gap = 1;
} else {
if (gap > 0) {
if (ans[i] == 0) {
ans[i] = gap;
} else {
ans[i] = Math.min(ans[i], gap);
}
gap++;
}
}
}
return ans;
}