[Solution] 821. Shortest Distance to a Character

Posted 2021-02-07 downstream-1998

• Difficulty: Easy

Problem

Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.

Example 1:

Input: S = "loveleetcode", C = ‘e‘
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]

Note:

1. S string length is in [1, 10000].
2. C is a single character, and guaranteed to be in string S.
3. All letters in S and C are lowercase.

Solution

找到 S 中第一个 C 的位置，然后从这里往两边扩展距离，然后找到下一个 C 的位置，以此类推，直到遍历完 S 中所有的 C。

实际上不难看出，位于 C 左边的字符，其距离计算只需进行一次（任意一个字符到 C 的距离，肯定比它到下一个 C 的距离更短），故向左扩展距离的时候就没必要扩展到字符串开头的位置。

public class Solution
{
    public int[] ShortestToChar(string S, char C)
    {
        int[] ret = new int[S.Length];
        int left = 0, right = S.Length, distance;
        int startIndex = S.IndexOf(C, left);

        Array.Fill(ret, 65535);

        while(startIndex != -1)
        {
            distance = 0;
            for(int i = startIndex; i >= left; i--)
            {
                if (ret[i] >= distance)
                    ret[i] = distance;
                distance++;
            }
            distance = 0;
            for(int i = startIndex; i < right; i++)
            {
                if (ret[i] >= distance)
                    ret[i] = distance;
                distance++;
            }
            left = startIndex;
            // 此处注意，查找下一个 C 的位置时，要排除掉当前的 C
            // 否则程序会陷入死循环
            startIndex = S.IndexOf(C, left + 1);
        }

        return ret;
    }
}

提交后在 LeetCode 榜单上发现另外一个解法

public class Solution
{
    public int[] ShortestToChar(string S, char C)
    {
        var len = S.Length;
        var retval = new int[len];
        var lastIdx = -len;
        for (int m = 0; m < len; m++)
        {
            if (S[m] == C) lastIdx = m;
            retval[m] = m - lastIdx;
        }

        lastIdx = len * 2;
        for (int m = len - 1; m >= 0; m--)
        {
            if (S[m] == C) lastIdx = m;
            retval[m] = Math.Min(lastIdx - m, retval[m]);
        }

        return retval;
    }
}