LeetCode每日一题(2405. Optimal Partition of String)

Given a string s, partition the string into one or more substrings such that the characters in each substring are unique. That is, no letter appears in a single substring more than once.

Return the minimum number of substrings in such a partition.

Note that each character should belong to exactly one substring in a partition.

Example 1:

Input: s = “abacaba”
Output: 4

Explanation:
Two possible partitions are (“a”,“ba”,“cab”,“a”) and (“ab”,“a”,“ca”,“ba”).
It can be shown that 4 is the minimum number of substrings needed.

Example 2:

Input: s = “ssssss”
Output: 6

Explanation:
The only valid partition is (“s”,“s”,“s”,“s”,“s”,“s”).

Constraints:

• 1 <= s.length <= 105
• s consists of only English lowercase letters.

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跟昨天的题很类似， 差不多的题， 不同的表述方式， 难度截然不同

基本谜底就在谜面上， 遍历字符串并同时维护一个 hashset 代表前面已经经历过的字符， 如果当前字符存在于 hashset 当中， 证明有重复的字符了， 我们要把前面的作为一个独立的 partition， 然后清空 hashset, 继续遍历

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use std::collections::HashSet;

impl Solution {
    pub fn partition_string(s: String) -> i32 {
        let mut curr = HashSet::new();
        let mut ans = 1;
        for c in s.chars() {
            if curr.contains(&c) {
                ans += 1;
                curr.clear();
            }
            curr.insert(c);
        }
        ans
    }
}