c++ - 地址 sanitizer :DEADLYSIGNAL

Question

我在leetcode的以下代码中使用`st.pop()时收到此错误:

AddressSanitizer:DEADLYSIGNAL
=================================================================
==31==ERROR: AddressSanitizer: SEGV on unknown address (pc 0x0000003a1925 bp 0x7ffd7b62dce0 sp 0x7ffd7b62dcd0 T0)
==31==The signal is caused by a READ memory access.
==31==Hint: this fault was caused by a dereference of a high value address (see register values below).  Dissassemble the provided pc to learn which register was used.
    #8 0x7f3d81a1e0b2  (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
AddressSanitizer can not provide additional info.
==31==ABORTING

代码如下:

class Solution {
public:
    string removeOuterParentheses(string S) {
        stack<char> st;
        int count=0;
        string ns;
        for(int i=0;i<S.size();i++)
        {
            if(S[i] == 40 && count++ > 0)
            {
                st.push(S[i]);
                ns+=S[i];
            }
            if(S[i] == 41 && count-- > 0)
            {
                st.pop();
                ns+=S[i];
            }
        }
        return ns;
    }
};

Answer 1

如果您有输入字符串"(some text)"，则条件S[i] == 40 && count++ > 0将始终评估为false。当S[i]=='('为true时，由于比较后count++ > 0递增，因此count为false。但是，count递增，这样第二个条件S[i] == ')' && count-- > 0最终将为true。然后您点击st.pop()。由于st为空，将导致问题。
另外，如果您修改了代码，例如通过编写S[i] == 40 && ++count > 0，不会删除括号，因为名称暗示了。