20. Valid Parentheses（stack）
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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.
public class Solution {
if (s == null)
return false;
Stack<Character> stack = new Stack<Character>();
char[] ch = s.toCharArray();
for (int i = 0; i < ch.length; i++) {
if ((ch[i] == ‘(‘) || (ch[i] == ‘[‘) || (ch[i] == ‘{‘))
stack.push(ch[i]);
else {
if (stack.isEmpty())
return false;
if (ch[i] - stack.pop() > 2)
return false;
}
}
return stack.isEmpty();
public boolean isValid(String s) {
if (s == null) //改进1：可以不用String.charat(i)
return false; //改进2： 先判断长度
//改进三： 不需要这么多if else ，符号的判断可以用ascii码
Stack<Character> stack = new Stack<Character>();
char[] ch = s.toCharArray();
for (int i = 0; i < ch.length; i++) {
if ((ch[i] == ‘(‘) || (ch[i] == ‘[‘) || (ch[i] == ‘{‘))
stack.push(ch[i]);
else {
if (ch[i] == ‘)‘) {
if (stack.isEmpty())
return false;
else if (stack.pop() != ‘(‘)
return false;
}
if (ch[i] == ‘]‘) {
if (stack.isEmpty())
return false;
if (stack.pop() != ‘[‘)
return false;
}
if (ch[i] == ‘}‘) {
if (stack.isEmpty())
return false;
if (stack.pop() != ‘{‘)
return false;
}
}
}
if (!stack.isEmpty())
return false;
else
return true;
}
}
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#Leetcode# 20.Valid Parentheses