Title
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
题目大意:就是走楼梯的最短路长，简单dp问题，第一次的就ac了，看solution的是倒着dp的，哈哈哈
//cpp:
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
vector<int> v(cost.size()+1);
fill(v.begin(), v.end(), 0);
for (int i = 0; i <= cost.size(); i++) {
if (i == 0)
v[i] = 0;
else if (i == 1)
v[i] == min(0, v[i - 1]);
else
v[i] = min(v[i - 2] + cost[i - 2], v[i - 1] + cost[i - 1]);
}
//cout << v[cost.size()];
return v[cost.size()];
}
};
class Solution {
public int minCostClimbingStairs(int[] cost) {
int f1 = 0, f2 = 0;
for (int i = cost.length - 1; i >= 0; --i) {
int f0 = cost[i] + Math.min(f1, f2);
f2 = f1;
f1 = f0;
}
return Math.min(f1, f2);
}
}