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LongestValidParentheses.java
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87 lines (75 loc) · 2.58 KB
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package Algorithms.string;
import java.util.Stack;
public class LongestValidParentheses {
public static void main(String[] strs) {
System.out.println(longestValidParentheses("(()()())"));
}
public static int longestValidParentheses(String s) {
if (s == null) {
return 0;
}
Stack<Integer> stk = new Stack<Integer>();
int sum = 0;
int tmp = 0;
int max = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '(') {
stk.push(i);
} else {
if (stk.isEmpty()) {
// 栈中没有'(',出现')', 则必须重置计算
sum = 0;
continue;
}
// count the temporary lenght:
// like: (()()()
// tmp = 2.
tmp = i - stk.pop() + 1;
if (stk.isEmpty()) {
// 有一套完整的括号集，可以加到前面的一整套括号集上
// () (()())
// 1 2 第二套括号集可以加过来
sum += tmp;
max = Math.max(sum, max);
} else {
// 也可能是一个未完成的括号集，比如：
// () (()() 在这里 ()() 是一个未完成的括号集，可以独立出来计算，作为
// 阶段性的结果
tmp = i - stk.peek();
max = Math.max(tmp, max);
}
}
}
return max;
}
public int longestValidParentheses2(String s) {
if (s == null) {
return 0;
}
int len = s.length();
Stack<Integer> stk = new Stack<Integer>();
int sum = 0;
int max = 0;
for (int i = 0; i < len; i++) {
char c = s.charAt(i);
if (c == '(') {
stk.push(i);
} else {
if (stk.isEmpty()) {
// The sequence is cut off.
sum = 0;
} else {
int tmp = i - stk.pop() + 1;
if (stk.isEmpty()) {
sum += tmp;
max = Math.max(max, sum);
} else {
max = Math.max(max, i - stk.peek());
}
}
}
}
return max;
}
}