Given a string containing just the characters '('
and ')'
, find the length of the longest valid (well-formed) parentheses substring.
Example 1:
Input: "(()"
Output: 2
Explanation: The longest valid parentheses substring is "()"
Example 2:
Input: ")()())
" Output: 4 Explanation: The longest valid parentheses substring is"()()"
大概意思是找寻最长有效括号的长度, 这是我的源代码
public int longestValidParentheses(String s) {
List<Integer> point = new LinkedList<Integer>();
int begin = 0, end = 0;
int oldBegin = 0, oldEnd = -2;
int arrzyIndex = -1;
for (int i = 0; i < s.length();) {
if (s.charAt(i) == '(' && i + 1 < s.length() && s.charAt(i + 1) == ')') {
begin = i;
end = i + 1;
while(begin - 1 >= 0 && s.charAt(begin - 1) == '(' && end + 1 < s.length() && s.charAt(end + 1) == ')') {
begin--; end++;
}
if(oldEnd + 1 == begin) {
begin = oldBegin;
while(begin - 1 >= 0 && s.charAt(begin - 1) == '(' && end + 1 < s.length() && s.charAt(end + 1) == ')') {
begin--; end++;
}
while(arrzyIndex > 2 && point.get(arrzyIndex - 2) + 1 == begin) {
begin = point.get(arrzyIndex - 3);
while(begin - 1 >= 0 && s.charAt(begin - 1) == '(' && end + 1 < s.length() && s.charAt(end + 1) == ')') {
begin--; end++;
}
point.remove(arrzyIndex);
point.remove(arrzyIndex - 1);
arrzyIndex -= 2;
}
point.set(arrzyIndex - 1, begin);
point.set(arrzyIndex, end);
} else {
point.add(begin);
point.add(end);
arrzyIndex += 2;
}
oldBegin = begin;
oldEnd = end;
i = end + 1;
} else {
i++;
}
}
int maxLength = -1;
for(int i = 0; i < point.size(); i += 2) {
if(point.get(i + 1) - point.get(i) > maxLength) {
maxLength = point.get(i + 1) - point.get(i);
}
}
return maxLength + 1;
}
这道题我想了很久,大概4天,由于近期心事太多的原因,最终采用最开始的方案。我的方案是这样的
1. 先找出所有的() (()) ..这样的括号
2.开始合并
3. 合并的时候注意 (A B) 这样的
4. 不断的重复 2、3即可
5. 找出最大的一组即可
看了看https://leetcode.com/problems/longest-valid-parentheses/solution/官方的解决方案,确实比我好。