746. Min Cost Climbing Stairs.md

746. Min Cost Climbing Stairs

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leetcode-cn Daily Challenge on December 21th, 2020.

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Difficulty : Easy

Related Topics : Array、Dynamic Programming

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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].

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Solution

• mine
  • Java Runtime: 1 ms, faster than 80.76%,Memory Usage: 39.2 MB, less than 57.14% of Java online submissions

    //O(N)time O(1)space
    public int minCostClimbingStairs(int[] cost) {
        int len = cost.length;
        for(int i = 2; i < len; i++ ){
            cost[i] = cost[i] + Math.min(cost[i - 1], cost[i - 2]);
        }
        return Math.min(cost[len - 1], cost[len - 2]);
    }
    

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• the leetcode's solution

  Dynamic Programming Runtime: 0 ms, faster than 100.00%, Memory Usage: 39 MB, less than 62.50% of Java online submissions

  public int minCostClimbingStairs(int[] cost) {
      int f1 = 0, f2 = 0;
      for (int i = cost.length - 1; i >= 0; --i) {
          int f0 = cost[i] + Math.min(f1, f2);
          f2 = f1;
          f1 = f0;
      }
      return Math.min(f1, f2);
  }