LeetCode - Min Cost Climbing Stairs

On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Example 1:
Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.
Example 2:
Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].
Note:
cost will have a length in the range [2, 1000].
Every cost[i] will be an integer in the range [0, 999].

 

Recursion Solution:

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        return Math.min(helper(cost, 0, 0), helper(cost, 1, 0));
    }
    
    public int helper(int[] cost, int index, int total) {
        if (index == cost.length - 2 || index == cost.length - 1) {
            return total + cost[index];
        }
        return Math.min(helper(cost, index+1, total+cost[index]), helper(cost, index+2, total+cost[index]));
    }
}

注意recurse 的方向，从开始recurse 能积累cache

Recursion With Memoization:

class Solution {
    public int minCostClimbingStairs(int[] cost) {
        int[] arr = new int[cost.length];
        return Math.min(helper(cost, cost.length-1, arr), helper(cost, cost.length-2, arr));
    }
    
    public int helper(int[] cost, int index, int[] arr) {
        if(index == 0 || index == 1) {
            return cost[index];
        }
        if (arr[index] > 0) {
            return arr[index];
        }
        arr[index] = Math.min(helper(cost, index-1, arr), helper(cost, index-2, arr)) + cost[index];
        return arr[index];
    }
}