Leetcode-Longest Valid Parentheses

LiBlog發表於2014-11-21
LeetCode

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.

Analysis:

At position i, if we get a ')', we find out the last '(', the current maxLen == i-lastPos+1. In addition, if s[lastPos-1] has maxLen[lastPos-1]>0, we need to add it to current maxLen, because the corresponding '(' for current position eliminate the obstacle between current max len sequence and the last max len sequence. For example, "())(())".

Solution 1:

We use a stack to store the index of each '('.

 1 public class Solution {
 2     public int longestValidParentheses(String s) {
 3         int n=0,m=0,curLen=0;
 4         List<Integer> pos = new ArrayList<Integer>();
 5         int[] maxLen = new int[s.length()];
 6         int finalRes = 0;
 7 
 8         for (int i=0;i<s.length();i++){
 9             char cur = s.charAt(i);
10             if (cur=='('){
11                 pos.add(i);
12                 maxLen[i]=0;
13             } else{
14                 if (pos.size()==0) maxLen[i]=0;
15                 else {
16                     int last = pos.get(pos.size()-1);
17                     pos.remove(pos.size()-1);
18                     curLen = i-last+1;
19                     if (last-1>=0) curLen += maxLen[last-1];
20                     maxLen[i] = curLen;
21                     if (finalRes<maxLen[i]) finalRes = maxLen[i];
22                 } 
23             }                   
24          }
25          return finalRes;
26     }
27 }

Solution 2:

We actually do not need to use a stack. The maxLen of the (i-1) position actually tell us the position of the last '('. 

 1 public class Solution {
 2     public int longestValidParentheses(String s) {
 3         if (s.length()==0) return 0;
 4         int n=0,m=0,curLen=0;       
 5         int[] maxLen = new int[s.length()];
 6         int finalRes = 0;
 7         maxLen[0] = 0;
 8         for (int i=1;i<s.length();i++){
 9             char cur = s.charAt(i);
10             if (cur=='('){               
11                 maxLen[i]=0;
12             } else{
13                 int lastLen = maxLen[i-1];
14                 int lastIndex = i-1-lastLen;
15                 
16                 if (lastIndex>=0 && s.charAt(lastIndex)=='('){
17                     maxLen[i] = lastLen+2;
18                     if (lastIndex-1>=0) maxLen[i] += maxLen[lastIndex-1];
19                     if (finalRes<maxLen[i]) finalRes = maxLen[i];
20                 }
21             }                   
22          }
23          return finalRes;
24     }
25 }

 

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