746.Min Cost Climbing Stairs
746.Min Cost Climbing Stairs
难度:Easy
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:
1. cost will have a length in the range [2, 1000].
2.Every cost[i] will be an integer in the range [0, 999].
直觀方法是逆序遞歸，但是計算量比較大，會顯示超時。
class Solution {
vector<int> c;
int minCost(int n)
{
if(n==1) return c[0];
if(n==2) return min(c[0],c[1]);
if(n==3) return min(c[0]+c[2],c[1]);
return min(minCost(n-1)+c[n-1], minCost(n-2)+c[n-2]);
}
public:
int minCostClimbingStairs(vector<int>& cost) {
for(auto i:cost)
c.push_back(i);
int len=cost.size();
return minCost(len);
}
};
另一種是順序DP，一邊循環。
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int n=cost.size();
if(n==0) return 0;
if(n==1) return cost[0];
if(n==2) return min(cost[0],cost[1]);
if(n==3) return min(cost[0]+cost[2],cost[1]);
vector<int> dp(n,0);
dp[0]=min(cost[0],cost[1]);
dp[1]=dp[0];
dp[2]=min(cost[1],cost[0]+cost[2]);
for(int i=3;i<n;i++)
dp[i]=min(dp[i-1]+cost[i],dp[i-2]+cost[i-1]);
return dp[n-1];
}
};
执行用时 :8 ms, 在所有 C++ 提交中击败了86.35%的用户 内存消耗 :8.7 MB, 在所有 C++ 提交中击败了84.38%的用户