Given a string containing just the characters ‘(‘
and ‘)‘
, find the length of the longest valid (well-formed) parentheses substring.
For "(()"
, the longest valid parentheses substring is "()"
, which has length = 2.
Another example is ")()())"
, where the longest valid parentheses substring is "()()"
, which has length = 4.
可以用 DP或者Stack来解。
栈 Stack:定义个start变量来记录合法括号串的起始位置,遍历字符串,如果遇到左括号,则将当前下标压入栈,如果遇到右括号,如果当前栈为空,则将下一个坐标位置记录到start,如果栈不为空,则将栈顶元素取出,此时若栈为空,则更新结果和i - start + 1中的较大值,否则更新结果和i - 栈顶元素中的较大值。
Java:
public static int longestValidParentheses(String s) { Stack<int[]> stack = new Stack<int[]>(); int result = 0; for(int i=0; i<=s.length()-1; i++){ char c = s.charAt(i); if(c==‘(‘){ int[] a = {i,0}; stack.push(a); }else{ if(stack.empty()||stack.peek()[1]==1){ int[] a = {i,1}; stack.push(a); }else{ stack.pop(); int currentLen=0; if(stack.empty()){ currentLen = i+1; }else{ currentLen = i-stack.peek()[0]; } result = Math.max(result, currentLen); } } } return result; }