LeetCode 316. Remove Duplicate Letters 栈顶

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本文介绍了一种去除字符串中重复字母的算法，确保结果在字典序中最小。通过使用栈和字典，该算法可以高效地处理问题，并提供了两种实现思路：一是利用栈和集合跟踪已存在的字符；二是从左至右选择最高优先级字符，递归处理剩余字符串。

Given a string which contains only lowercase letters, remove duplicate letters so that every letter appears once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.

Example 1:

Input: "bcabc"
Output: "abc"

Example 2:

Input: "cbacdcbc"
Output: "acdb"

Note: This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/

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能想到用栈这题就不难：

from collections import defaultdict
class Solution:
    def removeDuplicateLetters(self, s: str) -> str:
        sta,exist = [],set()
        dic = {ch:idx for idx,ch in enumerate(s)}
        for idx,ch in enumerate(s):
            if (ch not in exist):
                while (sta and sta[-1]>ch and dic[sta[-1]]>idx):
                    exist.discard(sta[-1])
                    sta.pop()
                sta.append(ch)
                exist.add(ch)
        return ''.join(sta)

另外一种思路是从左到右找优先级最高的字符，如果这个字符是最后一次出现，那么肯定应该放在这个位置。这个字符安置完，在原串中replace掉这个字符递归。

from collections import Counter

class Solution:
    def removeDuplicateLetters(self, s: str) -> str:

        c = Counter(s)
        pos = 0
        for i in range(len(s)):
            if s[i] < s[pos]: 
                pos = i
            c[s[i]] -=1
            if c[s[i]] == 0: 
                break
        return s[pos] + self.removeDuplicateLetters(s[pos:].replace(s[pos], "")) if s else ''