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".
來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/decode-string
1.字串中獲取數值
int getDigits() {
int ret = 0;
while (ptr < src.size() && isdigit(src[ptr])) {
ret = ret * 10 + src[ptr++] - '0';
}
return ret;
}
2.常用字符判斷函式
isdigit(cur) //當前字符是否為數字
isalpha(cur) //當前字符是否為字母
class Solution {
public:
string src;
int ptr = 0;
int getDigits() {
int ret = 0;
while (ptr < src.size() && isdigit(src[ptr])) {
ret = ret * 10 + src[ptr++] - '0';
}
return ret;
}
string getString() {
if (ptr == src.size() || src[ptr] == ']') {
return "";
}
char cur = src[ptr];
int repTime = 1;
string ret;
if (isdigit(cur)) {
repTime = getDigits();
++ptr;
string str = getString();
++ptr;
while (repTime--) ret += str;
}
else
if (isalpha(cur)) {
ret += src[ptr];
++ptr;
}
return ret + getString();
}
string decodeString(string s) {
src = https://www.cnblogs.com/xgbt/p/s;
ptr = 0;
return getString();
}
};
轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/49760.html
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