Leetcode 821. Shortest Distance to a Character

文章作者：Tyan
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1. Description

2. Solution

解析：Version 1，使用left，right分别记录当前字符的左右c的位置，左边没有为-1，右边没有为length。先初始化left，right，循环计算时，如果当前位置索引大于right，则对left，right进行更新，更新之后计算距离即可。

• Version 1

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        length = len(s)
        res = [0] * length
        left = -1
        right = 0
        while s[right] != c:
            right += 1
        for i in range(length):
            if right < i:
                left = right
                right += 1
                while right < length and s[right] != c:
                    right += 1
            if left < 0:
                res[i] = right - i
            elif right >= length:
                res[i] = i - left
            else:
                res[i] = min(i - left, right - i)
        return res

解析：Version 2，从左到右遍历数组，记录最左边字符c的位置，依次更新距离，从右往左遍历数组，记录最右边字符c的位置，依次更新。

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        length = len(s)
        res = [float('inf')] * length
        left = -1
        for i in range(length):
            if s[i] == c:
                left = i
            if left > -1:
                res[i] = min(res[i], i - left)
        right = length
        for i in range(length-1, -1, -1):
            if s[i] == c:
                right = i
            if right < length:
                res[i] = min(res[i], right - i)
        return res

Reference

1. https://leetcode.com/problems/shortest-distance-to-a-character/