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2405. Optimal Partition of String
給定一個字串s，我們想將 s 切成 n 個子字串，這些子字串不可以存在重複的字母，
求最少需要切成幾個子字串。
Example:
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.
Input: s = "ssssss"
Output: 6
Explanation:
The only valid partition is ("s","s","s","s","s","s").
思路：
1.res從1開始，因為最少存在一個字串。
2.統計當前切法的字母數量，若當前子字串加入新字元前發現重複，就開一個新的子字
串並讓res遞增，遍歷完一遍就完事了。
Java Code:
-----------------------------------------------
class Solution {
public int partitionString(String s) {
char[] count = new char[26];
int res = 1;
for (char c : s.toCharArray()) {
if (count[c - 'a'] == 1) {
count = new char[26];
res++;
}
count[c - 'a']++;
}
return res;
}
}
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※ 編輯: Rushia (122.100.75.86 臺灣), 04/04/2023 22:02:29
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04/04 22:16,
2年前
, 1F
04/04 22:16, 1F
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