746. Min Cost Climbing Stairs*
https://leetcode.com/problems/min-cost-climbing-stairs/
题目描述
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]
.
C++ 实现 1
动态规划. 用 dp[i]
表示达到 cost[i]
时所需要付出的最小代价, 则递推公式为:
dp[i] = std::min(dp[i - 1], dp[i - 2]) + cost[i]
实际上, 可以不用分配额外的空间 dp
, 直接修改原数组 cost
.
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int N = cost.size();
for (int i = 2; i < N; ++ i)
cost[i] += std::min(cost[i - 1], cost[i - 2]);
return std::min(cost[N - 2], cost[N - 1]);
}
};
C++ 实现 2
分配 dp
数组. 来自 LeetCode Submission.
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost)
{
int n=cost.size();
int dp[n];
dp[0]=cost[0];
dp[1]=cost[1];
for(int i=2;i<n;i++) {
dp[i]=min(dp[i-1],dp[i-2])+cost[i];
}
return min(dp[n-1],dp[n-2]);
}
};