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DAY 22
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30天 Leetcode解題之路系列 第 22

Day 22 - Shortest Distance to a Character

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廢話不多說開始今天的解題Day~


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

題目

首先先簡單的翻譯一下題目
給一個字串和字元,要先找出這個字元在字串的哪些位置上,最後回傳每個位置跟該字元的最短距離。

Think

作法大致上是這樣

  • 先找出cs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


Result

  • Python

  • C

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Day 23 - Keyboard Row
系列文
30天 Leetcode解題之路30

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

數數字囉4

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