stack--Longest Valid Parentheses
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<pre name="code" class="java">public class Solution { public int longestValidParentheses(String s) { Stack<Integer> stk = new Stack<Integer>(); int res = 0; char[] ch = s.toCharArray(); for(int i=0;i<ch.length;i++){ if(ch[i] == '(') stk.push(i); else{ if(!stk.isEmpty() && ch[stk.peek()] == '('){ stk.pop(); res = Math.max(stk.isEmpty()?i+1:i-stk.peek() ,res ); } else stk.push(i); } } return res; }}
stack存储上一个没匹配括号的index,遍历每个字符,发现栈顶的元素可以匹配当前括号,即为一对,否则再把当前括号入栈,作为未匹配的上一个index,以此类推。更新的时候如果stack为空,那么从头至尾所有的括号都匹配上,否则有效长度仅限i-stack.peek()。
Think about the following examples: "(()"
"()"
")())"
0 0
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