題目:
Given a string containing just the characters '('
, ')'
,
'{'
, '}'
, '['
and ']'
, determine if the input string is valid.
The brackets must close in the correct order, "()"
and "()[]{}"
are all valid but
"(]"
and "([)]"
are not.
思路與步驟:
如果是前半個括號,直接輸入;如果是後半個,則進行匹配。所以遇到兩個問題:
1. 用什麼數據結構來存儲括號;
2. 後半個匹配時的邏輯。
問題解決:
由於後半個匹配是要與最後一個輸入的來匹配(注意這裏的邏輯關係),所以可以直接用list,最經典的做法是用stack;同理第二個問題也得到解決,即讀入一個後半括號時,只有與上一個匹配纔是匹配。
一種巧妙解法:
上面提到的不管是用list還是stack,基本思路是一致的,只是數據結構的選擇不一樣。還有一種比較巧妙的思路,直接將前半個括號對應的後半個輸入stack,後面直接判斷是否相等即可。這樣可以使代碼更加簡潔。
編程實現:
這個問題的解決又複習了 Java 的 stack.
用List,更快:
public class Solution {
public boolean isValid(String s) {
//method-1: My own solution with list, faster than stack
char[] c = s.toCharArray();
if(s.length() == 0 || s.length()%2 == 1 || c[0] == ')' || c[0] == ']' || c[0] == '}') return false;
List<Character> list = new ArrayList<Character>();
for(int i=0; i<s.length(); i++){
if(c[i] == '(' || c[i] == '[' || c[i] == '{') list.add(c[i]);
else if(c[i] == ')'){
if(list.get(list.size()-1) == '(') list.remove(list.size()-1);
else return false;
} else if(c[i] == ']'){
if(list.get(list.size()-1) == '[') list.remove(list.size()-1);
else return false;
} else if(c[i] == '}'){
if(list.get(list.size()-1) == '{') list.remove(list.size()-1);
else return false;
}
}
if(list.size()==0) return true;
else return false;
/*//這一段效率不如上面那一段
for(int i=0; i<s.length(); i++){
if(c[i] == '(' || c[i] == '[' || c[i] == '{') list.add(c[i]);
else if(c[i] == ')' && list.get(list.size()-1) == '(') list.remove(list.size()-1);
else if(c[i] == ']' && list.get(list.size()-1) == '[') list.remove(list.size()-1);
else if(c[i] == '}' && list.get(list.size()-1) == '{') list.remove(list.size()-1);
else return false;
}
return list.size()==0 ? true : false;
*/
}
}
用stack:
public class Solution {
public boolean isValid(String s) {
//method-2: classic solution with stack
Stack<Character> stack = new Stack<Character>();
for (char c : s.toCharArray()) {
if (c == '(' || c == '[' || c == '{') stack.push(c);
else if(c == ')' && !stack.isEmpty() && stack.peek() == '(') stack.pop();
else if(c == ']' && !stack.isEmpty() && stack.peek() == '[') stack.pop();
else if(c == '}' && !stack.isEmpty() && stack.peek() == '{') stack.pop();
else return false;
}
return stack.isEmpty();
}
}
用stack,很巧妙的一種方法:
public class Solution {
public boolean isValid(String s) {
//method-3: an interesting solution with stack
Stack<Character> stack = new Stack<Character>();
for (char c : s.toCharArray()) {
if (c == '(') stack.push(')');
else if (c == '{') stack.push('}');
else if (c == '[') stack.push(']');
else if (stack.isEmpty() || stack.pop() != c) return false;
}
return stack.isEmpty();
}
}