- 斐波那契数
leetcode.com/problems/fi…
class Solution {
public int fib(int n) {
if (n <= 1) return n
int[] dp = new int[n + 1]
dp[0] = 0
dp[1] = 1
for (int index = 2
dp[index] = dp[index - 1] + dp[index - 2]
}
return dp[n]
}
}
- 爬楼梯
leetcode.com/problems/cl…
public int climbStairs(int n) {
int[] dp = new int[n + 1]
dp[0] = 1
dp[1] = 1
for (int i = 2
dp[i] = dp[i - 1] + dp[i - 2]
}
return dp[n]
}
- 使用最小花费爬楼梯
leetcode.com/problems/mi…
class Solution {
public int minCostClimbingStairs(int[] cost) {
int[] dp = new int[cost.length]
dp[0] = cost[0]
dp[1] = cost[1]
for (int i = 2
dp[i] = Math.min(dp[i - 1], dp[i - 2]) + cost[i]
}
//最后一步,如果是由倒数第二步爬,则最后一步的体力花费可以不用算
return Math.min(dp[cost.length - 1], dp[cost.length - 2])
}
}