Leetcode—20.有效的括号【简单】

2023每日刷题（二十七）

Leetcode—20.有效的括号

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C++实现代码

class Solution {
public:
    bool isValid(string s) {
        stack<char> arr;
        int len = s.size();
        if(len == 1) {
            return false;
        }
        for(int i = 0; i < len; i++) {
            if(s[i] == '(' || s[i] == '[' || s[i] == '{') {
                arr.push(s[i]);
            } else {
                if(arr.empty()) {
                    return false;
                }
                char t = s[i];
                char pairBracket = ' ';
                switch(t) {
                    case ')':
                        pairBracket = '(';
                        break;
                    case ']':
                        pairBracket = '[';
                        break;
                    case '}':
                        pairBracket = '{';
                        break;
                }
                if(arr.top() != pairBracket) {
                    return false;
                } else {
                    arr.pop();
                }
            }
        }
        if(arr.empty()) {
            return true;
        }
        return false;
    }
};

运行结果

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C语言实现代码

#define MAXSIZE 10003

typedef struct {
    char data[MAXSIZE];
    int top;
}SqStack;

void InitStack(SqStack *st) {
    st->top = -1;
}

bool stackEmpty(SqStack st) {
    return st.top == -1;
}

bool Push(SqStack *st, char e) {
    if(st->top == MAXSIZE - 1) {
        return false;
    }
    st->top++;
    st->data[st->top] = e;
    return true;
}

char Pop(SqStack *st) {
    char e = st->data[st->top];
    st->top--;
    return e;
}

char getTop(SqStack st) {
    return st.data[st.top];
}

bool isValid(char* s) {
    SqStack st;
    InitStack(&st);
    int n = strlen(s);
    for(int i = 0; i < n; i++) {
        if(s[i] == '(' || s[i] == '[' || s[i] == '{') {
            Push(&st, s[i]);
        } else {
            if(stackEmpty(st)) {
                return false;
            }
            char t = s[i];
            char pairBracket = ' ';
            switch(t) {
                case ')':
                    pairBracket = '(';
                    break;
                case ']':
                    pairBracket = '[';
                    break;
                case '}':
                    pairBracket = '{';
                    break;
            }
            if(getTop(st) != pairBracket) {
                return false;
            } else {
                Pop(&st);
            }
        }
    }
    if(stackEmpty(st)) {
        return true;
    }
    return false;
}