Insertion Sort List

Sort a linked list using insertion sort.

Bug free. Probably could do better.

Comments

Post a Comment

Popular posts from this blog

Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.  Try to solve it in linear time/space.  Return 0 if the array contains less than 2 elements. You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range. In order to solve this in linear time, we cannot use general sorting algorithm. Here we use bucket sort. First, we scan the array to find the min and the max values.   We can then create n - 1 buckets with equal size. Suppose we have n elements in total, then we have n - 2 elements except the min and the max. Scan the array again to put the n - 2 elements into these buckets. We should have at least one empty bucket. For each bucket, we record the min and the max for all the elements within that bucket.  Finally, we scan all the buckets, and maintain the maximum gap. Note that the maximum gap between the successive elements in the sorted...
Read more

House Robber

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.  Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police. class Solution { public: int rob(vector &num) { if (num.size() < 1) return 0; vector result; result.push_back(num[0]); for (int i = 1; i < num.size(); ++i) { int curr_max = 0; for (int j = 0; j < i - 1; ++j) { curr_max = max(curr_max, num[i] + result[j]); } curr_max = max(curr_max, result[i - 1]); curr_max = max(curr_max, num[i]); result.p...
Read more