746. Min Cost Climbing Stairs

1. Question

You are given an integer array cost where cost[i] is the cost of i^th step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

2. Examples

Example 1:

Input: cost = [10,15,20]
Output: 15
Explanation: You will start at index 1.

- Pay 15 and climb two steps to reach the top.
  The total cost is 15.

Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: You will start at index 0.

- Pay 1 and climb two steps to reach index 2.
- Pay 1 and climb two steps to reach index 4.
- Pay 1 and climb two steps to reach index 6.
- Pay 1 and climb one step to reach index 7.
- Pay 1 and climb two steps to reach index 9.
- Pay 1 and climb one step to reach the top.
  The total cost is 6.

3. Constraints

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/min-cost-climbing-stairs 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

5.1. dp

数组实现

class Solution {
  public int minCostClimbingStairs(int[] cost) {
    int prev = 0;
    int curr = 0;
    int next = 0;

    for (int i = 2; i <= cost.length; i++) {
      next = Math.min(curr + cost[i - 1], prev + cost[i - 2]);
      prev = curr;
      curr = next;
    }

    return next;
  }
}

5.2. dp算法的优化

滑动数组实现

class Solution {
  public int minCostClimbingStairs(int[] cost) {
    int prev = 0;
    int curr = 0;
    int next = 0;

    for (int i = 2; i <= cost.length; i++) {
      next = Math.min(curr + cost[i - 1], prev + cost[i - 2]);
      prev = curr;
      curr = next;
    }

    return next;
  }
}