821. 字符的最短距离

给定一个字符串 S 和一个字符 C。返回一个代表字符串 S 中每个字符到字符串 S 中的字符 C 的最短距离的数组。

示例 1:

输入: S = "loveleetcode", C = 'e'
输出: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
说明:

字符串 S 的长度范围为 [1, 10000]。
C 是一个单字符，且保证是字符串 S 里的字符。
S 和 C 中的所有字母均为小写字母。

 

class Solution:
    def shortestToChar(self, S: str, C: str) -> List[int]:
        n=len(S)
        res=[]
        c=[]
        for i in range(n):
            if S[i]==C:
                c.append(i)
        m=len(c)
        if m==1:
            if 0<c[0]<n-1:
                for i in range(c[0],-1,-1):
                    res.append(i)
                for i in range(n-c[0]-1):
                    res.append(i+1)
            elif c[0]==n-1:
                for i in range(c[0],-1,-1):
                    res.append(i)
            else:
                for i in range(n):
                    res.append(i)
        elif m==2:
            for i in range(c[0],-1,-1):
                res.append(i)
            if (c[1]-c[0])%2==1:
                for i in range(1,(c[1]-c[0])//2+1):
                    res.append(i)
                for i in range((c[1]-c[0])//2,-1,-1):
                    res.append(i)
            else:
                for i in range(1,(c[1]-c[0])//2+1):
                    res.append(i)
                for i in range((c[1]-c[0])//2-1,-1,-1):
                    res.append(i)
            for i in range(n-c[1]-1):
                res.append(i+1)
        else:
            for i in range(c[0],-1,-1):
                res.append(i)
            a=0
            b=1
            while b<=m-1:
                if (c[b]-c[a])%2==1:
                    for i in range(1,(c[b]-c[a])//2+1):
                        res.append(i)
                    for i in range((c[b]-c[a])//2,-1,-1):
                        res.append(i)
                else:
                    for i in range(1,(c[b]-c[a])//2+1):
                        res.append(i)
                    for i in range((c[b]-c[a])//2-1,-1,-1):
                        res.append(i)
                a+=1
                b+=1
            for i in range(n-c[-1]-1):
                res.append(i+1)
        return res

 

class Solution(object):
    def shortestToChar(self, S, C):
        pre = float('-inf')
        res = []
        for i, x in enumerate(S):
            if x == C: pre = i
            res.append(i - pre)

        pre = float('inf')
        for i in range(len(S) - 1, -1, -1):
            if S[i] == C: pre = i
            res[i] = min(res[i], pre - i)

        return res

 

class Solution:
    def shortestToChar(self, S: str, C: str) -> List[int]:
        return [min(S.find(C, i)-i, i-S[:i+1].rfind(C)) if S[:i+1].rfind(C) != -1 and S.find(C, i) != -1 else S.find(C, i)-i if S[:i+1].rfind(C) == -1 else i-S[:i+1].rfind(C) for i in range(len(S))]

 

class Solution:
    def shortestToChar(self, S: str, C: str) -> List[int]:
        pos = [i for i in range(len(S)) if S[i]==C]
        return([min(abs(x-i) for i in pos) for x in range(len(S))])

 

标签：pre,字符,str,res,len,range,短距离,821
来源： https://www.cnblogs.com/xxxsans/p/13752018.html