LeetCode-Longest Valid Parentheses

作者：disappearedgod
文章出处：/article/3730144.html
时间：2014-6-10

题目

Longest Valid Parentheses

Total Accepted: 9583 Total
Submissions: 50640My Submissions

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.

解法

非DP解法

public class Solution {
public int longestValidParentheses(String s) {
Queue<Integer> q = new LinkedList<Integer>();
int count = 0;
int result = 0;
if(s.length()<2)
return result;
for(int i = 0; i < s.length(); i++){
char c = s.charAt(i);
if(c=='(')
q.offer(++count);
if(c==')' && !q.isEmpty())
result = q.poll()*2;
}
return result;
}
}

Input:	"()(()"
Output:	4
Expected:	2

DP算法

这道题可以用一维动态规划逆向求解。假设输入括号表达式为String s，维护一个长度为s.length的一维数组dp[]，数组元素初始化为0。 dp[i]表示从s[i]到s[s.length - 1]最长的有效匹配括号子串长度。则存在如下关系：

dp[s.length - 1] = 0;
从i - 2 -> 0逆向求dp[]，并记录其最大值。若s[i] == '('，则在s中从i开始到s.length - 1计算s[i]的值。这个计算分为两步，通过dp[i + 1]进行的（注意dp[i + 1]已经在上一步求解）：

在s中寻找从i + 1开始的有效括号匹配子串长度，即dp[i + 1]，跳过这段有效的括号子串，查看下一个字符，其下标为j = i + 1 + dp[i + 1]。若j没有越界，并且s[j] == ‘)’，则s[i ... j]为有效括号匹配，dp[i] =dp[i + 1] + 2。
在求得了s[i ... j]的有效匹配长度之后，若j + 1没有越界，则dp[i]的值还要加上从j + 1开始的最长有效匹配，即dp[j + 1]。

public class Solution {
public int longestValidParentheses(String s) {
int result = 0;
int len = s.length();
int[] dp = new int[len];// store the location of ')' macthed
char c = 0;
//initialize dp with 0 in JAVA
for(int i = len -2 ; i >= 0 ; i-- ){
c = s.charAt(i);
if(c=='('){
int j = i + 1 + dp[i+1];//find the location of ')', which matches '('
if(j < len &&  s.charAt(j) == ')'){// match = no overflow and get the ')'
dp[i] = dp[i+1] +2;// (())
int flag = 0;
if(j + 1< len ){ // Adding the result of the next'(',for example()()
flag = dp[j +1];
}
dp[i] += flag;
}
}
result = Math.max(result, dp[i]);
}
return result;
}
}

参考 http://blog.csdn.net/abcbc/article/details/8826782