DAY 22
0

# 821. Shortest Distance to a Character

## Question

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

### Example1

``````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.
``````

### Example2

``````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`.

## 解題

### Think

• 先找出`c``s`的哪些位置上(這邊存在`indices`的list中)，再來用`Binary Search`去找`s`的每一個index應該插在`indices`的哪個位置上。
• C再補上，先睡啦。

### Code

#### Python

``````class Solution:
def shortestToChar(self, s: str, c: str) -> List[int]:
indices = [index for index, value in enumerate(s) if value == c]
# print(indices)
# print(len(s))

ans = []

for index in range(len(s)):
if index in indices:
ans.append(0)
continue

# binary Search
low = 0
upper = len(indices)-1
mid = 0

while low <= upper:
mid = int(round( ((low + upper) / 2), 0))

if indices[mid] < index:
low = mid + 1
elif indices[mid] > index:
upper = mid - 1
else:
ans.append(abs(indices[mid]-index))

# If it doesn't contain in indices, we need to find where it would be inserted.
if indices[mid] < index:
if mid+1 <= len(indices)-1:
ans.append(min(abs(indices[mid]-index), abs(indices[mid+1]-index)))
else:
ans.append(abs(indices[mid]-index))

else:
if mid-1 >= 0:
ans.append(min(abs(indices[mid-1]-index), abs(indices[mid]-index)))
else:
ans.append(abs(indices[mid]-index))

return ans
``````

#### C

``````
``````

• Python

• ## C

30天 Leetcode解題之路30

### 1 則留言

0
soft_soft
iT邦新手 5 級 ‧ 2021-10-08 20:38:49