Min-Cost-Climbing-Stairs

发表于 | 更新于：

题目地址

LeetCode#746 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:

1
2
3
Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top.

Example 2:

1
2
3
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].

解题思路

  这种全局最优解的题要么递归要么动态规划，这道题递归超时，动态规划可以直接解。

解题代码【.CPP】

1
2
3
4
5
6
7
8
9
10
11
12
13
class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int size = static_cast<int>(cost.size());
        vector<int> dp(size,0);
        dp[0] = cost[0];
        dp[1] = cost[1];
        for (int i = 2; i < size; ++i) {
            dp[i] = cost[i] + min(dp[i-1] , dp[i-2]);
        }
        return min(dp[size-1],dp[size-2]);
    }
};
0%