Leetcode 1653. Minimum Deletions to Make String Balanced [Python]

设置只压入字母‘b’的stack，遇到a，则看stack中是否有字母‘b’，有的话，则需要删除，则每次出stack，res+1.而在任意位置之后，无论stack有没有东西，只要之后遍历到的是‘b’而没有了‘a’或者一个字母都没有了，那么就说明遍历到的字母都满足a在前，b在后了。总之，保证遍历到‘b’后不再遇到‘a’，遇到‘a’则把其之前的‘b’删除。那么如果例子是‘aabba’怎么办？那么在遇到最后一个a的时候，stack需要pop一个b out。而逻辑上，其实是删除了a，同理，如果是‘aabbaa’,那么我逻辑上可以是删除了两个stack中的‘b’，也可以说是删除了最后的两个‘a’。而如果是‘aabbaaa’ ，那么说明我删除掉不符合顺序的‘b’操作次数更少，逻辑上就是删除了两个‘b’。

class Solution:def minimumDeletions(self, s: str) -> int:stack = []res = 0for idx,c in enumerate(s):if c == 'b':stack.append(c)elif stack:stack.pop()res += 1return res

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