按摩师 - 简单

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C++

topic: 面试题 17.16. 按摩师 - 力扣（LeetCode）

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Inspect the topic first.

And It just remainds me another topic, which is stocks problems.

买卖股票的最佳时机 - 合集-优快云博客https://blog.youkuaiyun.com/ElseWhereR/article/details/144608599?spm=1001.2014.3001.5501

By the way, do you know the king of massage?

Topic like this is to define the states of the each day. In the day i, receive the appointment or reject the appointment. Define dp[i] is the maxmum work time. 

for the i appointment, two choices, receive or reject.

1. accepts the i-th appointment, but then she can't accept the i-1th appointment, because the total time is dp[i-2] + nums[i].
2. doesn't accept the i-th appointment, then the total time is dp[i-1].

So the transfer equation is 

dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]).

int dp[0] = nums[1];

int dp[1] = max(nums[1], nums[2])

for ecample:

Suppose `nums = [1, 2, 3, 4, 5]`.

- `n = 5`

- `dp[0] = 1`

- `dp[1] = max(1, 2) = 2`

- `dp[2] = max(2, 1 + 3) = max(2, 4) = 4`

- `dp[3] = max(4, 2 + 4) = max(4, 6) = 6`

- `dp[4] = max(6, 4 + 5) = max(6, 9) = 9`

So the maximum profit is 9, which seems correct. She can take appointments 1 (1), 3 (3), and 5 (5), totaling 9.

class Solution {
public:
    int massage(vector<int>& nums) {

        // always initilize something here
        int n = nums.size();

        vector<int> dp(n, 0);
        dp[0] = nums[1];
        dp[1] = max(nums[1], nums[2]);
        
        for (int i = 2; i < n; i++){
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        
        return dp[n];
    }
};

The code looks beautiful and runs error.

I forget about the boundary conditions. 

if n = 0; return 0;

if n = 1; return nums[0];

class Solution {
public:
    int massage(vector<int>& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        
        vector<int> dp(n, 0);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);
        
        for (int i = 2; i < n; i++) {
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        
        return dp[n - 1];
    }
};