[LeetCode] 1358. Number of Substrings Containing All Three Characters

Given a string s consisting only of characters a, b and c.

Return the number of substrings containing at least one occurrence of all these characters a, b and c.

Example 1:

Input: s = "abcabc"
Output: 10
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "abc", "abca", "abcab", "abcabc", "bca", "bcab", "bcabc", "cab", "cabc" and "abc" (again). 

Example 2:

Input: s = "aaacb"
Output: 3
Explanation: The substrings containing at least one occurrence of the characters a, b and c are "aaacb", "aacb" and "acb".

Example 3:

Input: s = "abc"
Output: 1

Constraints:

• 3 <= s.length <= 5 x 10^4
• s only consists of a, b or c characters.

包含所有三种字符的子字符串数目。

给你一个字符串 s ，它只包含三种字符 a, b 和 c 。请你返回 a，b 和 c 都 至少 出现过一次的子字符串数目。

思路是滑动窗口。这道题的窗口卡的是一个子区间满足子区间内unique key的个数是3。同时因为input里只有abc三个字母，所以当你找到第一个子串的终点 end 的时候，子串[start, s.length - 1] 其实都是满足题意的子串，所以此时可以尝试移动 start 指针，只有count == 3的时候，我们才能移动 start 指针。这道题依然可以使用76题的模板。

时间O(n)

空间O(1) - letter数组可以忽略不计

Java实现

 1 class Solution {
 2     public int numberOfSubstrings(String s) {
 3         int len = s.length();
 4         int[] letter = new int[3];
 5         int count = 0;
 6         int res = 0;
 7         int start = 0;
 8         int end = 0;
 9         while (end < s.length()) {
10             char c1 = s.charAt(end);
11             if (letter[c1 - 'a']++ == 0) {
12                 count++;
13             }
14             while (count == 3) {
15                 res += len - end;
16                 char c2 = s.charAt(start);
17                 if (letter[c2 - 'a']-- == 1) {
18                     count--;
19                 }
20                 start++;
21             }
22             end++;
23         }
24         return res;
25     }
26 }

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来源： https://www.cnblogs.com/cnoodle/p/14182375.html