Given a nested list of integers represented as a string, implement a parser to deserialize it.
Each element is either an integer, or a list -- whose elements may also be integers or other lists.
Note: You may assume that the string is well-formed:
- String is non-empty.
- String does not contain white spaces.
- String contains only digits
0-9,[,-,,].
Example 1:
Given s = "324", You should return a NestedInteger object which contains a single integer 324.
Example 2:
Given s = "[123,[456,[789]]]", Return a NestedInteger object containing a nested list with 2 elements: 1. An integer containing value 123.
2. A nested list containing two elements:
i. An integer containing value 456.
ii. A nested list with one element:
a. An integer containing value 789.
这道题让我们实现一个迷你解析器用来把一个字符串解析成NestInteger类，关于这个嵌套链表类的题我们之前做过三道，Nested List Weight Sum II，Flatten Nested List Iterator，和Nested List Weight Sum。应该对这个类并不陌生了，我们可以先用递归来做，思路是，首先判断s是否为空，为空直接返回，不为空的话看首字符是否为'['，不是的话说明s为一个整数，我们直接返回结果。如果首字符是'['，且s长度小于等于2，说明没有内容，直接返回结果。反之如果s长度大于2，我们从i=1开始遍历，我们需要一个变量start来记录某一层的其实位置，用cnt来记录跟其实位置是否为同一深度，cnt=0表示同一深度，由于中间每段都是由逗号隔开，所以当我们判断当cnt为0，且当前字符是逗号或者已经到字符串末尾了，我们把start到当前位置之间的字符串取出来递归调用函数，把返回结果加入res中，然后start更新为i+1。如果遇到'['，计数器cnt自增1，若遇到']'，计数器cnt自减1。参见代码如下：
解法一：
class Solution {
public:
NestedInteger deserialize(string s) {
if (s.empty()) return NestedInteger();
if (s[] != '[') return NestedInteger(stoi(s));
if (s.size() <= ) return NestedInteger();
NestedInteger res;
int start = , cnt = ;
for (int i = ; i < s.size(); ++i) {
if (cnt == && (s[i] == ',' || i == s.size() - )) {
res.add(deserialize(s.substr(start, i - start)));
start = i + ;
} else if (s[i] == '[') ++cnt;
else if (s[i] == ']') --cnt;
}
return res;
}
};
class Solution {
public:
NestedInteger deserialize(string s) {
if (s.empty()) return NestedInteger();
if (s[] != '[') return NestedInteger(stoi(s));
stack<NestedInteger> st;
int start = ;
for (int i = ; i < s.size(); ++i) {
if (s[i] == '[') {
st.push(NestedInteger());
start = i + ;
} else if (s[i] == ',' || s[i] == ']') {
if (i > start) {
st.top().add(NestedInteger(stoi(s.substr(start, i - start))));
}
start = i + ;
if (s[i] == ']') {
if (st.size() > ) {
NestedInteger t = st.top(); st.pop();
st.top().add(t);
}
}
}
}
return st.top();
}
};
还有一种方法是利用C++ STL中的字符串流处理类istringstream，我们需要对几个函数有些了解，比如clear()是重置字符流中的字符串，get()是获得下一个字符，peek()是返回首字符，>>num是读取出合法的整数，如果无法读取出整数，需要调用clear()来重置字符串，否则调用get()会出错。思路跟上面的递归解法相同，参见代码如下：
解法三：
class Solution {
public:
NestedInteger deserialize(string s) {
istringstream in(s);
return deserialize(in);
}
NestedInteger deserialize(istringstream& in) {
int num;
if (in >> num) return NestedInteger(num);
in.clear();
in.get();
NestedInteger list;
while (in.peek() != ']') {
list.add(deserialize(in));
if (in.peek() == ',') {
in.get();
}
}
in.get();
return list;
}
};
类似题目：
参考资料：
https://discuss.leetcode.com/topic/54258/python-c-solutions/3
https://discuss.leetcode.com/topic/54341/iterative-c-using-stack
https://discuss.leetcode.com/topic/54277/short-java-recursive-solution