算法刷题day28|动态规划：509. 斐波那契数、70. 爬楼梯、746. 使用最小花费爬楼梯

dp套路

1. 确定dp数组（dp table）以及下标的含义
2. 确定递推公式
3. dp数组如何初始化
4. 确定遍历顺序
5. 举例推导dp数组

509. 斐波那契数

class Solution {
public:
    int fib(int n) {
        if (n <= 1) return n;
        int dp[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        for (int i = 2; i <= n; i++){
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
};

70. 爬楼梯

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 2) return n;
        int dp[n + 1];
        dp[0] = 0;
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++){
            dp[i] = dp[i - 2] + dp[i - 1];
        }
        return dp[n];
    }
};

746. 使用最小花费爬楼梯

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int dp[cost.size() + 1];
        dp[0] = dp[1] = 0;
        for (int i = 2; i <= cost.size(); i++){
            dp[i] = min(cost[i - 1] + dp[i - 1], cost[i - 2] + dp[i - 2]);
        }
        return dp[cost.size()];
    }
};