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leetcode 2166 设计位集直接用bitset做

booksmg2014 2022-02-11 阅读 49

参考:https://leetcode-cn.com/problems/design-bitset/
用C++中stl中自带的bitset做,注意的是bitset得固定申请大小,不能动态在函数中申请,所以在最开始先申请好,之后在求all,one,count的时候要注意处理,即最后一段数是不会被修改的全0或者全1,用标记计算即可。

class Bitset {
public:
    bitset<100001>s;
    int n,other;
    Bitset(int size) {
        n=size;
        other=100001-size;
    }
    
    void fix(int idx) {
        s[idx]=1;
    }
    
    void unfix(int idx) {
        s[idx]=0;
    }
    
    void flip() {
        s=~s;
    }
    
    bool all() {
        if(s[100000]==1)return s.count()==s.size();
        else return s.count()==n;
    }
    
    bool one() {
        if(s[100000]==0)return s.any();
        else return s.count()>other;
    }
    
    int count() {
        if(s[100000]==0)return s.count();
        else return s.count()-other;
    }
    
    string toString() {
        string res;
        for(int i=0;i<n;i++)res+=s[i]+'0';
        return res;
    }
};

/**
 * Your Bitset object will be instantiated and called as such:
 * Bitset* obj = new Bitset(size);
 * obj->fix(idx);
 * obj->unfix(idx);
 * obj->flip();
 * bool param_4 = obj->all();
 * bool param_5 = obj->one();
 * int param_6 = obj->count();
 * string param_7 = obj->toString();
 */
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