Regular Expression Matching
Problem description:
Implement regular expression matching with support for ‘.‘ and ‘*‘.
‘.‘ Matches any single character.
‘*‘ Matches zero or more of the preceding element.
The matching should cover the entire input string (not partial).
The function prototype should be:
bool isMatch(const char *s, const char *p)
Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true
解题思路：
不是显式DP问题（保存一张表），用Recursion的思路可解。
1）若p[j+1] != ‘*‘，那么判断p[j]和s[i]的结果就可以了。
p[j]==s[i]或者p[j]==‘.‘，到当前位置是匹配的，继续向下判断；不满足条件，说明当前位置不匹配。
2）若p[j+1]==‘*‘，那么需要循环判断p[j+2]和s[i...n]是否匹配.
代码如下：
1 class Solution {
2 public:
3 bool isMatch(const char *s, const char *p) {
4 if(*p == 0)
5 return *s == 0;
6
7 if(*(p+1) != ‘*‘)
8 {
9 if(*s != 0 && (*p == *s || *p == ‘.‘))
10 return isMatch(s+1, p+1);
11 else
12 return false;
13 }
14 else
15 {
16 while(*s != 0 && (*p == *s || *p == ‘.‘))
17 {
18 if(isMatch(s, p+2))
19 return true;
20
21 s++;
22 }
23 return isMatch(s, p+2);
24 }
25 }
26 };
Longest Valid Parentheses
Problem description:
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.
解题思路：
不是用DP解决的，用stack解决。使用pair<char, int>保存字符和位置。
1）如果s[i]==‘(‘，则压入栈；
2）如果s[i]==‘)‘，则判断st.top()是否是‘(‘，如果是则弹出栈；否则压入栈。
代码如下：
1 class Solution {
2 public:
3 int longestValidParentheses(string s) {
4 stack<pair<char,int> > st;
5 int len = s.length();
6
7 if(len <= 1)
8 return 0;
9
10 pair<char,int> start(‘)‘,-1);
11 st.push(start);
12 int res = 0;
13 for(int i = 0; i < len; ++i)
14 {
15 if(s[i] == ‘(‘)
16 {
17 pair<char,int> node(s[i],i);
18 st.push(node);
19 }
20 else
21 {
22 pair<char,int> top = st.top();
23 if(top.first == ‘(‘)
24 {
25 st.pop();
26 res = max(res, i-st.top().second);
27 }
28 else
29 {
30 pair<char,int> node(s[i],i);
31 st.push(node);
32 }
33
34 }
35 }
36
37 return res;
38 }
39 };