Given an encoded string, return its decoded string.
The encoding rule is: k[encoded_string]
, where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.
You may assume that the input string is always valid; No extra white spaces, square brackets are well-formed, etc.
Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, k. For example, there won't be input like 3a
or 2[4]
.
Examples:
s = "3[a]2[bc]", return "aaabcbc".
s = "3[a2[c]]", return "accaccacc".
s = "2[abc]3[cd]ef", return "abcabccdcdcdef".
题解:
这题很像224和227题。
用一个栈保存前面所有字符结果和前一个数字大小。
题解:
class Solution {
public:
string decodeString(string s) {
int n = s.length();
string res = "";
stack<string> str;
int num = 0;
for (int i = 0; i < n; i++) {
if (s[i] >= '0' && s[i] <= '9') {
num *= 10;
num += s[i] - '0';
}
else if (s[i] == '[') {
str.push(res);
str.push(to_string(num));
num = 0;
res.clear();
}
else if (s[i] == ']') {
int k = stoi(str.top());
str.pop();
string tmp = str.top();
str.pop();
for (int j = 0; j < k; j++) {
tmp += res;
}
res = tmp;
}
else {
res += s[i];
}
}
return res;
}
};