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  • 表达式的逆波兰式转化模板

    表达式的逆波兰式
    ##使用方法 输入合法的表达式,加减乘除,可以带括号,例如: `a+b*(c-d)-e/f` ##模板 ``` #include #include #include #include //#include using namespace std; #define mem(x,num) memset(x,num,sizeof(x)) const int N = 1e3 + 5; const int defaultSize = 100; int lson[N], rson[N]; int nc = 0;

    char op[N];
    void Init() {
    mem(lson, 0);
    mem(rson, 0);
    mem(op, 0);
    nc = 0;
    }
    int build_Tree(char s[], int l, int r) { // Build the expression tree
    int tag1 = -1, tag2 = -1, p = 0;// tag1 stand for add&sub is_exist,tag2 for muti&devide
    int u;
    if (r - l == 1) {// this is only one point in this interval,build it
    u = ++nc;
    lson[u] = rson[u] = 0;
    op[u]=s[l];
    return u;// return self num to fa Node
    }
    for (int i(l); i < r; i++) {
    switch (s[i])// s[i]
    {
    case '(':p++; break;
    case ')':p--; break;
    case '+':case '-':if(!p)tag1 = i; break;// if operator in (),continue
    case '*':case '/':if(!p)tag2 = i; break;
    }
    }
    if (tag1 < 0)tag1 = tag2;// no + or - outside
    if (tag1 < 0)return build_Tree(s, l + 1, r - 1); // no * or / outside mean all operator and num in ()
    u = ++nc;
    lson[u] = build_Tree(s, l, tag1);
    rson[u] = build_Tree(s, tag1 + 1, r);
    op[u] = s[tag1];
    return u;
    }
    void Post(int rt) {// Postorder
    if (lson[rt] != 0)Post(lson[rt]);
    if (rson[rt] != 0)Post(rson[rt]);
    cout << op[rt] << ' ';
    }
    int main() {
    char s[defaultSize];
    while (cin>>s) {
    Init();
    int n=0;
    build_Tree(s, 0,strlen(s));
    Post(1);
    cout << endl;
    }
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/Titordong/p/10472687.html
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