[LeetCode] 746. Min Cost Climbing Stairs

轉自LeetCode

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Note: Given n will be a positive integer.

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps

Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step

<Solution>

想法如下

• 用 recursive 會 TLE，必須優化，所以考慮 DP 來優化

• 用 dp[i] 來表示在 i 所要花費的最少 cost，公式是 dp[i] = cost[i] + min(dp[i-1], dp[i-2])

code 如下

Java

	class Solution {
	public int minCostClimbingStairs(int[] cost) {
	int oneStepBefore = cost[1], twoStepBefore = cost[0];
	int curStepCost;
	for(int i = 2; i < cost.length; i++) {
	curStepCost = cost[i] + Math.min(oneStepBefore, twoStepBefore);
	twoStepBefore = oneStepBefore;
	oneStepBefore = curStepCost;
	}
	return Math.min(oneStepBefore, twoStepBefore);
	}
	}

view raw minCostClimbingStairs.java hosted with ❤ by GitHub

kotlin

	class Solution {
	fun minCostClimbingStairs(cost: IntArray): Int {
	var oneStepBeforeCost = cost[1]
	var twoStepBeforeCose = cost[0]
	var currentCost = 0
	for(i in 2 until cost.size) {
	currentCost = cost[i]+ Math.min(oneStepBeforeCost, twoStepBeforeCose)
	twoStepBeforeCose = oneStepBeforeCost
	oneStepBeforeCost = currentCost
	}
	return Math.min(oneStepBeforeCost, twoStepBeforeCose)
	}
	}

view raw min_cost_climbing_stairs.kt hosted with ❤ by GitHub