LeetCode解题报告（289）-- 821. Shortest Distance to a Character

Problem

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 1:

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) = 3.
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.

Example 2:

Input: s = "aaab", c = "b"
Output: [3,2,1,0]

Constraints:

• 1 <= s.length <= 104
• s[i] and c are lowercase English letters.
• It is guaranteed that c occurs at least once in s.

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Analysis

  题目给出一个字符串以及一个字符c，问字符串上每个位置距离该字符的最短距离。在字符串中除了c以外，其他字符距离最近的c要么在它的左边，要么在它的右边，所以我们只需要从左往右遍历一遍，再从右往左遍历一遍，就能找到答案了。

  从左往右遍历的时候，除了c以外，每个位置的距离比前一个i - 1递增1；同理，从右往左遍历时，除了c之外的位置的距离比前一个i + 1 递增1。第一遍计算的时候不需要取最小值，第二遍计算的时候需要和第一遍的结果进行比较，取极小值。

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Solution

  这里初始化有个细节需要注意，因为是求最小值，所以初始化的时候把值设置为数组的长度即可。

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Code

class Solution {
public:
    vector<int> shortestToChar(string s, char c) {
        int size = s.size();
        vector<int> result(size, size);
        for (int i = 0; i < size; i++) {
            if (s[i] == c) {
                result[i] = 0;
            } else if (i > 0) {
                result[i] = result[i - 1] + 1;
            }
        }
        
        for (int i = size - 2; i >= 0; i--) {
            result[i] = min(result[i], result[i + 1] + 1);
        }
        return result;
    }
};

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Summary

  这道题目的分享到这里，谢谢您的支持！