Leetcode

# 20. Valid Parentheses

by hooni posted Apr 25, 2020
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Given a string containing just the characters `'('``')'``'{'``'}'``'['` and `']'`, determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

```Input: "()"
Output: true
```

Example 2:

```Input: "()[]{}"
Output: true
```

Example 3:

```Input: "(]"
Output: false
```

Example 4:

```Input: "([)]"
Output: false
```

Example 5:

```Input: "{[]}"
Output: true```

```/*
class Solution {
public boolean isValid(String s) {
Map<Character, Integer> hmap = new HashMap<>();
Map<Character, Character> pairs = new HashMap<>();

pairs.put(')', '(');
pairs.put('}', '{');
pairs.put(']', '[');

for(int i=0; i<s.length(); i++){
Character ch = s.charAt(i);

if(ch == '(' || ch == '{' || ch == '['){
hmap.put(ch, hmap.getOrDefault(ch,0)+1);
}else if(ch == ')' || ch == '}' || ch == ']'){
Character pair = pairs.get(ch);

if(hmap.containsKey(pair) == false){
return false;
}

hmap.put(pair, hmap.get(pair)-1);
System.out.println(hmap);
System.out.println("-> "+pair);

if(hmap.get(pair) == 0){
hmap.remove(pair);
}
}
}

return hmap.isEmpty();
}
}
*/
class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
Map<Character, Character> pairs = new HashMap<>();

pairs.put(')', '(');
pairs.put('}', '{');
pairs.put(']', '[');

for(int i=0; i<s.length(); i++){
Character ch = s.charAt(i);

if(ch == '(' || ch == '{' || ch == '['){
stack.push(ch);
}else if(ch == ')' || ch == '}' || ch == ']'){
if(stack.empty()){
return false;
}

if(stack.peek() == pairs.get(ch)){
stack.pop();
}else{
return false;
}
}
}

return stack.isEmpty();
}
}```

```class Solution {
public boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
Map<Character, Character> pairs = new HashMap<>();

pairs.put(')', '(');
pairs.put('}', '{');
pairs.put(']', '[');

for(int i=0; i<s.length(); i++){
Character ch = s.charAt(i);

//if(ch == '(' || ch == '{' || ch == '['){
if(pairs.containsValue(ch)){
stack.push(ch);
//}else if(ch == ')' || ch == '}' || ch == ']'){
}else if(pairs.containsKey(ch)){
if(stack.empty()){
return false;
}

if(stack.peek() == pairs.get(ch)){
stack.pop();
}else{
return false;
}
}
}

return stack.isEmpty();
}
}```

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