[Leetcode] 32. longest effective brackets

Keywords: DP, dynamic programming, dynamic rules.

In recent brush DP column topic, which is among a question.

Given that contains only '('and ')'string, to find the longest sub-string comprises an effective parentheses.

Sample1

输入: "(()"
输出: 2
解释: 最长有效括号子串为 "()"

Sample2

输入: ")()())"
输出: 4
解释: 最长有效括号子串为 "()()"

For DP Well, above all, the need to abstract state function:

dp[i] 表示：以 S[i] 结尾的，最长有效括号串的长度。

Then the state transition equation:

如果 s.len == 0 or s.len == 1，那么返回 0 .
否则：
    if s[i]=='(' then dp[i]=0
    if s[i]==')' then:
        if s[i-1]='(' then dp[i] = dp[i-2] + 2 （数组下标是否越界，即 i>=2? ）
        if s[i-1]=')' then:
            if s[i-dp[i-1]-1] == '(' then dp[i] = dp[i-dp[i-1]-2] + dp[i-1] + 2 (是否越界？)
            if s[i-dp[i-1]-1] == ')' then dp[i] = 0

The first ifstatement indicates that: the form ....(of such a string, is not necessarily valid.

The second ifstatement said: shaped like ....)such a string, so we need to consider s[i-1]:

s[i-1]='(': String shaped like ...(), obviously, correspond to the following index:
```
i-2   i-1    i
 x     (     )       
```
Obviously, dp[i]the value should be dp[i-2] + 2.

s[i-1]='): String shaped like ...)), obviously, correspond to the following index:

?          i-1    i
x   (...    )     )

Now consider s[i-1]the matching positions of the left parenthesis , s[i-1]=')'which is legitimate parentheses string length dp[i-1], the '('position should be:

i - 1 - （dp[i-1] - 1) = i - dp[i-1]

In other words, s[i] = )the matching left parenthesis position should be: i - dp[i-1] - 1.

i-dp[i-1]-1  i-dp[i-1]        i-1    i
    x            (      ...    )     )

If x = s[i-dp[i-1]-1] == ), then: dp[i] = dp[i-dp[i-1]-2] + dp[i-1] + 2(because "I did not draw out" in front may still be effective in parentheses string)

But if the i-dp[i-1]-1symbol of this position is not (it? This shows that with s[i]parenthesis at the end of the pass is not legitimate, namely: d[i] = 0.

Complete code:

Require special attention Array Subscript out of range of the problems, once the bounds, the foregoing description is not a valid string bracket.

/*
    DP解法：
    dp[i] 表示：以s[i]结尾的，最长有效子串
    那么：
        if s[i]=='(' then dp[i] = 0
        if s[i]==')':
            if (s[i-1]=='(') then dp[i] = dp[i-2]+2
            if (s[i-1]==')') then dp[i] = dp[i-dp[i-1]-2] + dp[i-1] + 2

 */
#include "leetcode.h"
#include <stack>
class Solution
{
public:
    int longestValidParentheses(string s)
    {
        int len = s.length();
        if (len == 0 || len == 1)
            return 0;
        vector<int> dp(len, 0);
        for (int i = 1; i < len; i++)
        {
            if (s[i] == ')')
            {
                if (s[i - 1] == '(')
                {
                    if (i >= 2)
                        dp[i] = dp[i - 2] + 2;
                    else
                        dp[i] = 2;
                }
                else if (s[i - 1] == ')')
                {
                    int midlen = dp[i - 1];
                    if (i >= (midlen + 1))
                    {
                        char c = s[i - midlen - 1];
                        if (c == '(')
                            dp[i] = dp[i - 1] + 2 + (i >= (midlen + 2) ? dp[i - midlen - 2] : 0);
                        else
                            dp[i] = 0;
                    }
                    else
                    {
                        dp[i] = 0;
                    }

                }
            }
        }
        for (int x : dp)
            cout << x << ' ';
        cout << endl;
        int result = -1;
        for (int x : dp)
            result = max(result, x);
        return result;
    }
};

int main()
{
    string s[] = {"())", "(()", ")()())"};
    Solution sol;
    for (int i = 0; i < 3; i++)
        cout << sol.longestValidParentheses(s[i]) << "\n"
             << endl;
}

[Leetcode] 32. longest effective brackets

[Leetcode] 32. longest effective brackets

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