• Problem
  • Example 1
  • Example 2
  • Constraints
• Approach 1: Dynamic Programming
  • Algorithm
  • Implementation
  • Complexity Analysis

You are given an integer array cost where cost[i] is the cost of ith 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.

• 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.

• 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.

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

可創建一個 slice dp, dp[i] 表示爬到第 i 階所需要花費的最少步數, 為前一階或是前兩階的最小步數加上當前階層所需步數, 即:

dp[i] = min(dp[i-2], dp[i-1]) + cost[i]

func minCostClimbingStairs(cost []int) int {
	l := len(cost)
	dp := make([]int, l, l)
	dp[0] = cost[0]
	dp[1] = cost[1]

	for i := 2; i < l; i++ {
		dp[i] = min(dp[i-2], dp[i-1]) + cost[i]
	}
	return min(dp[l-1], dp[l-2])
}

func min(i, j int) int {
	if i < j {
		return i
	} else {
		return j
	}
}

• Time complexity: O(n)

  • Where n is the length of given slice.
  • Runtime: 0 ms, faster than 100.00% of Go online submissions for Unique Paths.

• Space complexity: $O(m\ast n)$

  • Where n is the length of given slice.
  • Memory Usage: 3.1 MB, less than 43.33% of Go online submissions for Min Cost Climbing Stairs.