如果当前的字符为数位,解析出一个数字并进栈如果当前的字符为字母或者左括号,直接进栈如果当前的字符为右括号,开始出栈,一直到左括号出栈,出栈序列反转后拼接一个字符串,此时取出栈顶的数字,就是这个字符串应该出现的次数,根据这个次数和字符串构造出新的字符串并进栈
class Solution {
public:
string getDigits(string &s, size_t &ptr) {
string ret = "";
while (isdigit(s[ptr])) {
ret.push_back(s[ptr++]);
}
return ret;
}
string getString(vector <string> &v) {
string ret;
for (const auto &s: v) {
ret += s;
}
return ret;
}
string decodeString(string s) {
vector <string> stk;
size_t ptr = 0;
while (ptr < s.size()) {
char cur = s[ptr];
if (isdigit(cur)) {
string digits = getDigits(s, ptr);
stk.push_back(digits);
} else if (isalpha(cur) || cur == '[') {
stk.push_back(string(1, s[ptr++]));
} else {
++ptr;
vector <string> sub;
while (stk.back() != "[") {
sub.push_back(stk.back());
stk.pop_back();
}
reverse(sub.begin(), sub.end());
stk.pop_back();
int repTime = stoi(stk.back());
stk.pop_back();
string t, o = getString(sub);
while (repTime--) t += o;
stk.push_back(t);
}
}
return getString(stk);
}
};