# Leetcode 746. Min Cost Climbing Stairs
###### tags: `Leetcode`

題目
You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

Example 1:

Input: cost = [10,15,20]
Output: 15
Explanation: You will start at index 1.
- Pay 15 and climb two steps to reach the top.
The total cost is 15.
Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1]
Output: 6
Explanation: You will start at index 0.
- Pay 1 and climb two steps to reach index 2.
- Pay 1 and climb two steps to reach index 4.
- Pay 1 and climb two steps to reach index 6.
- Pay 1 and climb one step to reach index 7.
- Pay 1 and climb two steps to reach index 9.
- Pay 1 and climb one step to reach the top.
The total cost is 6.

解法:
1.計算當下step所需要的cost = 當下cost加上較小的前一個或前前一個的cost
2.若要知道當下step最小的cost，就是看當下的cost跟前一step的cost，因為要走到現在這個step，有兩種方式，一種就是直接走到目標的step，一種就是走到目標的前一step

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int min(int a, int b)
{
    if(a>b)
        return b;
    else 
        return a;
}
int minCostClimbingStairs(int* cost, int costSize){
    for(int i = 2; i < costSize; i++)
    {
        cost[i] += min(cost[i-1], cost[i-2]);
    }
    return min(cost[costSize-1], cost[costSize-2]);
}