LeetCode 2399. Check Distances Between Same Letters【哈希表,字符串】简单

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You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26.

Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, … , 'z' -> 25).

In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.

Return true if s is a well-spaced string, otherwise return false.

Example 1:

Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.

Example 2:

Input: s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: false
Explanation:
- 'a' appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.

Constraints:

• 2 <= s.length <= 52
• s consists only of lowercase English letters.
• Each letter appears in s exactly twice.
• distance.length == 26
• 0 <= distance[i] <= 50

题意：字符串 $s$ 中每个小写字母都出现了两次，看相同的两个小写字母之间的字符数是否等于 distance[i] ，i 为该小写字母在字母表中出现的顺序。

解法 哈希表

只要能想办法把初始状态与遍历过一次的状态区分开来，就很简单：

class Solution {
public:bool checkDistances(string s, vector<int>& distance) {int d[26]; memset(d, -1, sizeof(d));for (int i = 0, n = s.size(); i < n; ++i) {int v = s[i] - 'a';if (d[v] == -1) d[v] = i;else if (i - d[v] - 1 != distance[v]) return false;}return true;}
};