[Leetcode] 818. Race Car 解题报告

题目：

Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)

Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).

When you get an instruction "A", your car does the following: position += speed, speed *= 2.

When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)

For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.

Now for some target position, say the length of the shortest sequence of instructions to get there.

Example 1:
Input: 
target = 3
Output: 2
Explanation: 
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.

Example 2:
Input: 
target = 6
Output: 5
Explanation: 
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.

Note:

1 <= target <= 10000.

思路：

采用动态规划的思路，定义dp[target]表示行驶长度为target的距离所需要的最小指示个数。可以证明有如下两种可能性：

1）如果target刚好是可以由“AAA...AR”（一共n步）达到，也就是说前面一路加速，最后再减速到-1，那么这种走法就是最优选择；

2）如果不满足条件1），那么最优解就有多种可能性了：a）第一次冲过target的时候进行‘R’操作，然后反向再接近target。此时我们已经走了n+1步，并且和target的距离已经减少到了(1 << n) - 1 - target，所以我们可以递归地调用rececar函数；b）前面先走m步（可以通过递归调用racecar函数得到最优值），然后再采用1）的策略。最终的答案就是a）和b）的各种可能性的最小值。

为了达到最优的运行效率，我们采用dp + memorization的策略：每当需要计算一个子问题时，首先查表看看该子问题是否已经被计算过了；如果是，则直接返回结果；否则我们再进行计算，并且将结果保存下来，以备后面的计算再次使用。

代码：

class Solution {
public:
    int racecar(int target) {
        if (dp[target] > 0) {                       // already calculated before
            return dp[target];
        }
        int n = floor(log2(target)) + 1, res;
        if (1 << n == target + 1) {                 // perfect solution: just use someting like "AAA...AR"
            dp[target] = n;
        }
        else {                                      // need to go back using the same strategy
            dp[target] = racecar((1 << n) - 1 - target) + n + 1;
            for (int m = 0; m < n - 1; ++m) {
                dp[target] = min(dp[target], racecar(target - (1 << (n - 1)) + (1 << m)) + n + m + 1);
            }
        }
        return dp[target];
    }
private:
    int dp[10001];
};

[Leetcode] 818. Race Car 解题报告

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