我在网上发现的该函数的时间复杂度为O（n）吗？

Question

I am trying to do this problem on LeetCode , and the following is the solution I found online:

 // use quick select var findKthLargest = function(nums, k) { var smaller = []; var larger = []; var pivot = nums[parseInt(nums.length/2)]; var pivotCnt = 0; for (var i = 0; i < nums.length; i++) { var num = nums[i]; if(num > pivot) { larger.push(num); } else if(num < pivot) { smaller.push(num); } else { pivotCnt++; } } if (k <= larger.length) { // if larger includes k return findKthLargest(larger, k); } else if (k - larger.length - pivotCnt <= 0) { // k is part of pivot return pivot; } else { // if smaller inclues k return findKthLargest(smaller, k - larger.length - pivotCnt); } }; 

Now I believe this is an O(n) solution, because worst case is that we iterate through the entire array, but I'm not certain.

Answer 1

Your function appears to be using some sort of divide and conquer approach. For each call, it makes a single O(n) pass over the input array, bucketing values larger than, and smaller than, a certain pivot, into two separate arrays. Then, it makes a recursive call on the appropriate subarray. In an average case, it will divide the size of the input array by two, until a recursive call is made on just an array of size one, which is the base case.

I would estimate that the running time of this function is O(n*lgn) , which is typical for a divide and conquer algorithms. Each call does O(n) work, and there would typically be O(lgn) number of recursive calls.