leetcode 746. Min Cost Climbing Stairs(easy understanding dp solution)

leetcode 746. Min Cost Climbing Stairs(easy understanding dp solution)

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].

solution

top要么是 top - 1 上去的，要么是 top -2 上去的

故动态规划的状态转移方程为：

dp[i] = min{dp[i-1]+cost[i-1], dp[i -2]+cost[i-2]}

可以从第0阶开始，也可以从第1阶开始，故动态规划处理两次即可

class Solution {
public:
	int minCostClimbingStairs(vector<int>& cost) {
		int n = cost.size();
		int dp[n + 1];
		dp[0] = 0;
		dp[1] = cost[0];
		for (int i = 2; i <= n; i++)
			dp[i] = (dp[i - 1] + cost[i - 1]) < (dp[i - 2] + cost[i - 2]) ? (dp[i - 1] + cost[i - 1]) : (dp[i - 2] + cost[i - 2]);

		int dp1[n + 1];
		dp1[0] = 0;
		dp1[1] = cost[1];
		for (int i = 2; i <= n - 1; i++)
			dp1[i] = (dp1[i - 1] + cost[i]) < (dp1[i - 2] + cost[i - 1]) ? (dp1[i - 1] + cost[i]) : (dp1[i - 2] + cost[i - 1]);

		if (dp1[n - 1] < dp[n])
			return dp1[n - 1];
		else
			return dp[n];

	}
};

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