Number of cycles of a permutation

J, 4 bytes

#@C.

This assumes that the permutation is 0-based. It uses the builtin C. which given a list representing a direct permutation outputs a list of cycles. Then # composed @ on that returns the number of cycles in that list.

Try it here.

JavaScript, 99 98 bytes

This solution assumes the array and its values are zero-indexed (e.g. [2, 1, 0]).

f=a=>{h={},i=c=0;while(i<a.length){s=i;while(!h[i]){h[i]=1;i=a[i]}c++;i=s;while(h[++i]);}return c}

Explanation

// assumes the array is valid and zero-indexed
var findCycles = (array) => {
    var hash = {};  // remembers visited nodes
    var index = 0;  // current node
    var count = 0;  // number of cycles
    var start;      // starting node of cycle
    
    // loop until all nodes visited
    while(index < array.length) {
        start = index;  // cache starting node
        
        // loop until found previously visited node
        while(!hash[index]) {
            hash[index] = 1;    // mark node as visited
            index = array[index];   // get next node
        }
        count++;    // increment number of cycles
        
        index = start + 1;  // assume next node is right after
        
        // loop until found unvisited node
        while(hash[index]) {
            index++;    // get next node
        }
    }
    
    return count;   // return number of cycles
};

Python, 77 69 67 bytes

f=lambda p,i=1:i and0 **p[i-1]+f(p[:i-1]+[0]+p[i:],p[i-1]or max(p))