【POJ HDOJ leetcode】括号匹配合法性及最长括号匹配

/*

1. string parenthesis
给出一个由()组成的字符串判断合法性,如()合法, (,  (((不合法.

2. 给出一串()字符串,求出最长连续括号匹配的长度及其位置
*/

 

#include <iostream>
#include <stdio.h>
#include <stack>
using namespace std;
class Solution {
public:
    bool isValid(const string& s) {
        if (s == "") {
            return true;
        }
        stack<char> stk;
        size_t size = s.size();
        for (size_t i = 0; i < size; i++) {
            if (s[i] == '(') {
                stk.push(s[i]);
            } else {
                if (stk.empty()) return false;
                stk.pop();
            }
        }
        return stk.size() == 0;
    }
};

pair<int,int> NumOfMatch(const char* str)
{
    if (str == NULL) return {0, 0};
    const char* p = str;
    int nLeft = 0;      // 待匹配的左括号的个数
    int nMatchedPre = 0;// 上一个匹配子串的已匹配括号的对数
    int nMatchedCur = 0;// 当前子串的已匹配括号的对数
    int maxi = -1, maxmatch = 0;
    const char* front = p;
    while (*p != '\0') {
        if (*p == '(') {
            ++nLeft;
        } else if (*p == ')') {
            if (nLeft == 0) {
                maxmatch = max(nMatchedCur, maxmatch);
                nMatchedPre = nMatchedCur;
                nMatchedCur = 0;
            } else {
                nMatchedCur++;
                nLeft--;
                if (nMatchedCur > maxmatch) {
                    maxi = (p - front);
                    maxmatch = nMatchedCur;
                }
            }
        }
        p++;
    }
    maxi -= maxmatch * 2 - 1;
    maxmatch *= 2;
    return make_pair(maxi, maxmatch);
}
 
int cnt = 0;
int total = 0;
void expect_test(bool a, bool b) {
    total++;
    if (a == b) {
        cnt++;
        return;
    }
    printf("expect %d actual %d\n", a, b);
}

void expect_test_int(int a, int b) {
    total++;
    if (a == b) {
        cnt++;
        return;
    }
    printf("expect %d actual %d\n", a, b);
}
typedef pair<int,int> pii;
void expect_test_pair(pii a, pii b) {
    total++;
    if (a.first == b.first && a.second == b.second) {
        cnt++;
        return;
    }
    printf("expect {%d,%d} actual {%d,%d}\n",
    a.first, a.second, b.first, b.second);
}
int main() {
    Solution sol;
    expect_test(true, sol.isValid("()"));
    expect_test(true, sol.isValid("(())"));
    expect_test(true, sol.isValid("()()"));
    expect_test(false, sol.isValid(")"));
    expect_test_pair({0, 2}, NumOfMatch("()))"));
    expect_test_pair({0, 4}, NumOfMatch("(()))"));
    expect_test_pair({0, 4}, NumOfMatch("(())"));
    expect_test_pair({0, 0}, NumOfMatch(""));
    expect_test_pair({4, 4}, NumOfMatch("())((())"));
    printf("%d/%d\n", cnt, total);
    return 0;
}

 

来源：oschina

链接：https://my.oschina.net/u/4327596/blog/3426043