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Min Cost Climbing Stairs
On a staircase, the
i-th step has some non-negative costcost[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.
costwill have a length in the range[2, 1000]. - 2.Every
cost[i]will be an integer in the range[0, 999].
朴素的动态规划DP - Bottom Up
这里比较明显的状态是用
dp[i]表示到达第i个step时所需要的cost；递推关系也容易得到，状态转移方程是
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i]
相当于
i - 1, i - 2 位置的cost加上当前i位置的cost初始状态 dp[0] = cost[0], dp[1] = cost[1] 分别代表从0，或者1出发
最终结果则是取距离top为1或者2 steps中的较小值
Math.min(dp[ len - 1], dp[ len - 2])DP - O(n) space, O(n) time - (13ms 36.68%)
class Solution {
public int minCostClimbingStairs(int[] cost) {
int len = cost.length;
int[] climbCost = new int[len];
climbCost[0] = cost[0];
climbCost[1] = cost[1];
for (int i = 2; i < len; i++) {
climbCost[i] = Math.min(climbCost[i - 1], climbCost[i - 2]) + cost[i];
}
return Math.min(climbCost[len - 1], climbCost[len - 2]);
}
}
DP - O(1) space, O(n) time
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);
}
}
- Time Complexity:O(N) where N is the length of
cost. - Space Complexity:O(1), the space used by
f1, f2.
Last modified 3yr ago