M
1520800825
tags: Divide and Conquer, Stack, DFS
给一个expression string. 里面包括数字, 字母, 括号. 其中数字代表括号里面的内容重复几次.
括号里面可以是String, 也可能是expression.
目的: 把expression展开成一个正常的String.
#### Stack, Iteratively
- Process inner item first: last come, first serve, use stack.
- Record number globally and only use it when '[' is met.
- Stack存 [ ] 里面的内容, detect 括号开头结尾: 结尾时process inner string
- 有很多需要注意的细节才能做对:
- Stack 也可以用, 每个地方要注意 cast. 存进去的需要是Object: String, Integer
- 几个 type check: instanceof String, Character.isDigit(x), Integer.valueOf(int num)
- 出结果时候: `sb.insert(0, stack.pop())`
#### DFS
- Bottom->up: find deepest inner string first and expand from inside of `[ ]`
- 与Stack时需要考虑的一些function类似. 特别之处: **检查`[ ]`的结尾**
- 因为DFS时候, 括号里的substring会被保留着进入下一个level, 所以我们在base level要keep track of substring.
- 用int paren 来track 括号的开合, 当paren再次==0的时候 找到closure ']'
- 其他时候, 都要继续 append to substring
```
/*
Given an encoded string, return it's 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".
*/
public class Solution {
public String decodeString(String s) {
if (s == null || s.length() == 0) {
return "";
}
Stack stack = new Stack<>();
int number = 0;
for (char c : s.toCharArray()) {
if (Character.isDigit(c)) {
number = number * 10 + (c - '0');
} else if (c == '[') {
stack.push(Integer.valueOf(number));
number = 0;
} else if (c == ']') {
String str = popStack(stack);
int num = (Integer) stack.pop();
for (int i = 0; i < num; i++) stack.push(str);
} else {
stack.push(String.valueOf(c));
}
}
return popStack(stack);
}
private String popStack(Stack stack) {
StringBuffer sb = new StringBuffer();
while (!stack.isEmpty() && (stack.peek() instanceof String)) {
sb.insert(0, stack.pop());
}
return sb.toString();
}
}
/*
Given an expression s includes numbers, letters and brackets.
Number represents the number of repetitions inside the brackets
(can be a string or another expression)．
Please expand expression to be a string.
Example
s = abc3[a] return abcaaa
s = 3[abc] return abcabcabc
s = 4[ac]dy, return acacacacdy
s = 3[2[ad]3[pf]]xyz, return adadpfpfpfadadpfpfpfadadpfpfpfxyz
*/
/*
Thoughts:
Can DFS. Also, use stack to hold all inner bucket.
In the original string, block of string is separated by #[ ].
Can use stack to hold 2 piece of information:
1. # where the str repeats, and it also marks the start of the string
2. Store each char of the string itself
3. Once flatten the inner str, repeat the str # times and add back to stack
As we iterate over the string
- detect '[' to determine the time to add start marker #.
- detect ']' when it's time to retrieve the # and str, save results
Important: use Stack to hold generic Object!
*/
public class Solution {
public String expressionExpand(String s) {
if (s == null || s.length() == 0) {
return "";
}
Stack stack = new Stack<>();
int number = 0;
for (char c : s.toCharArray()) {
if (Character.isDigit(c)) {
number = number * 10 + (c - '0');
} else if (c == '[') {
stack.push(Integer.valueOf(number));
number = 0;
} else if (c == ']') {
String str = popStack(stack);
Integer num = (Integer) stack.pop();
for (int i = 0; i < num; i++) {
stack.push(str);
}
} else {
stack.push(String.valueOf(c));
}
}
return popStack(stack);
}
private String popStack(Stack stack) {
StringBuffer sb = new StringBuffer();
Stack buffer = new Stack<>();
while (!stack.isEmpty() && (stack.peek() instanceof String)) {
buffer.push((String) stack.pop());
}
while (!buffer.isEmpty()) {
sb.append(buffer.pop());
}
return sb.toString();
}
}
/*
Thoughts:
DFS, process inner string each time.
Use "#[" as detection.
Important:
1. Finding the closure of ']' is tricky: need to track the parentheses
2. Until we find the closure ']', keep track of the un-flatten substring for dfs to use
*/
public class Solution {
/**
* @param s: an expression includes numbers, letters and brackets
* @return: a string
*/
public String expressionExpand(String s) {
if (s == null || s.length() == 0) {
return "";
}
StringBuffer sb = new StringBuffer();
String substring = "";
int number = 0;
int paren = 0; // parentheses tracker
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '[') {
if (paren > 0) { // if paren == 0, it indicates the outside 1st '[', no need to record
substring += c;
}
paren++;
} else if (c == ']') {
paren--;
if (paren == 0) {
String innerString = expressionExpand(substring);
for (int num = 0; num < number; num++) {
sb.append(innerString);
}
number = 0;
substring = "";
} else {
substring += c;
}
} else if (Character.isDigit(c)) {
if (paren == 0) {
number = number * 10 + (c - '0');
} else {
substring += c;
}
} else {
if (paren == 0) {
sb.append(String.valueOf(c));
} else {
substring += c;
}
}
}
return sb.toString();
}
}
```