746. Min Cost Climbing Stairs

Problem

Tags: Array, Dynamic Programming

You are given an integer array cost where cost[i] is the cost of i^th 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.

Constraints:

• 2 <= cost.length <= 1000
• 0 <= cost[i] <= 999

Code

C

// 746. Min Cost Climbing Stairs (10/7/53956)
// Runtime: 4 ms (79.53%) Memory: 5.70 MB (91.73%) 

int minCostClimbingStairs(int cost[], int costSize){
    // 差一步「之前」的總 cost 跟差兩步「之前」的總 cost，起始 0 意義是第零階的 cost
    int one = 0,
        two = 0;
    
    // 滾上去，記得要滾到 costSize
    for (int i = 2; i <= costSize; i++) {
        // 每次往上一層，差一步「之前」的總 cost 會變成差兩步「之前」的總 cost
        int new_two = one;
        // 而新的差一步「之前」的總 cost 則會是原本差一步「之前」的總 cost + 差一步的 cost 與原本差兩步「之前」的總 cost + 差兩步的 cost，兩者較小值
        one = one + cost[i - 1] < two + cost[i - 2] ? one + cost[i - 1] : two + cost[i - 2];
        two = new_two;
    }

    // 最後輸出的其實是到 n + 1 階差一步「之前」的總 cost
    return one;
}

JS

// 746. Min Cost Climbing Stairs (9/28/53402)
// Runtime: 84 ms (49.42%) Memory: 41.04 MB (94.53%) 

/**
 * @param {number[]} cost
 * @return {number}
 */
var minCostClimbingStairs = function(cost) {
    let one = 0, two = 0;
    for(let i = 2; i <= cost.length; i++) {
        [one, two] = [(one + cost[i - 1]) < (two + cost[i - 2]) ? (one + cost[i - 1]) : (two + cost[i - 2]), one]
    }
    return one;
};