678. Valid Parenthesis String

Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules:
Any left parenthesis 

'('

 must have a corresponding right parenthesis 

')'

.
Any right parenthesis 

')'

 must have a corresponding left parenthesis 

'('

.
Left parenthesis 

'('

 must go before the corresponding right parenthesis 

')'

.

'*'

 could be treated as a single right parenthesis 

')'

 or
a single left parenthesis 

'('

 or an empty string.
An empty string is also valid.

Example 1:

Input: "()"
Output: True

Example 2:

Input: "(*)"
Output: True

Example 3:

Input: "(*))"
Output: True

Note:

The string size will be in the range [1, 100].

思路：自己想的办法比较烦，类似于维持一个左括号的计数，该值不能小于0

package l678;

class Solution {
public boolean checkValidString(String s) {
int left = 0, star4left = 0, star = 0, setPoint = 0;
char[] cs = s.toCharArray();

for(int i=0; i<cs.length; i++) {
char c = cs[i];
if(c == '(')	left++;
else if(c=='*')star++;
else if(c==')')left--;

// 少left，就从*号里面取出一个来表示，而且这些星号还不能表示成')'
if(left == -1) {
star4left += star;	// 不能表示右括号了
if(star4left == 0)	return false;
left = 0;
star4left --;
star = 0;
setPoint = i+1;
} else if(left == 0) {
star4left += star;	// 不能表示右括号了
star = 0;
setPoint = i+1;
}
}

if(left == 0)	return true;
// 如果右括号有多，还要再遍历一遍
int t = 0, starUseCnt = 0;
for(int i=setPoint; i<cs.length; i++) {
if(cs[i] == '(')	t++;
else if(cs[i]==')')t--;
else if(t>0) {
t--;
starUseCnt ++; 	// 先把*认为是)，记录这样替换的次数
}

if(t == -1) {
if(starUseCnt == 0) return false;
starUseCnt--;	// 之前*可能替换成)，替换过头了，判断下
t = 0;
}
}

return t <= 0;
}
}

有人是：维护左括号的lower bound 和 upper bound

the idea is to similar to the solution of without '*'. keep a lower bound of '(' counts and an upper bound of '(' count.

public boolean checkValidString(String s) {
int low = 0;
int high = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == '(') {
low++;
high++;
} else if (s.charAt(i) == ')') {
if (low > 0) {
low--;
}
high--;
} else {
if (low > 0) {
low--;
}
high++;
}
if (high < 0) {
return false;
}
}
return low == 0;
}

也有很naive的递归

How to check valid parenthesis w/ only ( and )? Easy. Count each char from left to right. When we see (, count++; when we see ) count--; if count < 0, it is invalid () is more than (); At last, count should == 0.
This problem added *. The easiest way is to try 3 possible ways when we see it. Return true if one of them is valid.
class Solution {
public boolean checkValidString(String s) {
return check(s, 0, 0);
}

private boolean check(String s, int start, int count) {
if (count < 0) return false;

for (int i = start; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '(') {
count++;
}
else if (c == ')') {
if (count <= 0) return false;
count--;
}
else if (c == '*') {
return check(s, i + 1, count + 1) || check(s, i + 1, count - 1) || check(s, i + 1, count);
}
}

return count == 0;
}
}