您将如何使用for循环实现O（C ^ n）时间复杂度函数？

Question

The only thing I can think of uses Math.pow() to calculate n before the loop and seems like a cop out.
C=2 in this example:

var oCN = function (n)
{
    var j = Math.pow(2, n);
    for (var i = 0; i <= j; i++) {
          console.log('cactus');  
    };
    return;
}

Answer 1

A recursive algorithm like this one would grow exponentially with respect to n :

var oCN = function (C, n)
{
    if (n < 1) { console.log('cactus'); return; }
    for (var i = 0; i < C; i++) {
        oCN(C, n-1);
    }
}

The number of logged cacti is in the case of this algorithm is exactly C^n , which is of course O(C^n) . Eg. oCN(2, 4) logs cactus 16 times, oCN(2, 5) logs it 32 times, etc.

You won't typically see a lot of day-to-day algorithms that work in exponential time. If something is being done in exponential time there is a good chance the input size or constants are such that the big-O complexity doesn't matter very much at all (or at least, the implementer doesn't care very much), and so you probably won't be sitting around analyzing the complexity. Most normal stuff programmers want to do can either be done with a polynomial time algorithm right off the bat or with some fudging and sacrificing.