[LeetCode] 1266. Minimum Time Visiting All Points 訪問所有點的最小時間

On a 2D plane, there are n points with integer coordinates points[i] = [xi, yi]. Return *the minimum time in seconds to visit all the points in the order given by *points.

You can move according to these rules:

• In 1 second, you can either:
  • move vertically by one unit,
  • move horizontally by one unit, or
  • move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second).
• You have to visit the points in the same order as they appear in the array.
• You are allowed to pass through points that appear later in the order, but these do not count as visits.

Example 1:

Input: points = [[1,1],[3,4],[-1,0]]
Output: 7
Explanation: One optimal path is [1,1] -> [2,2] -> [3,3] -> [3,4] -> [2,3] -> [1,2] -> [0,1] -> [-1,0]
Time from [1,1] to [3,4] = 3 seconds
Time from [3,4] to [-1,0] = 4 seconds
Total time = 7 seconds

Example 2:

Input: points = [[3,2],[-2,2]]
Output: 5

Constraints:

• points.length == n
• 1 <= n <= 100
• points[i].length == 2
• -1000 <= points[i][0], points[i][1] <= 1000

這道題給了一堆二維平面上的點，讓按照順序去連接點，這里的連線不但可以走水平和豎直，還能走斜線，都算作一步，問按順序連上所有的點需要多少步。這里讓按順序連點也就簡單了不少，也算對得起 Easy 的身價，只要分析出兩個點之間的最小步數怎么算就可以解題了。題目中說可以水平，豎直，和對角線走，那么就按照題目中的例子來分析吧，若仔細觀察可以發現，假如兩個點的橫縱坐標的差值相等的話（都是絕對值），那么只要純走對角線就行了，比如 [3,4] 和 [-1,0] 這兩個點，若差值不相等的話，則同時要走對角線和橫縱邊，比如 [1,1] 和 [3,4] 這兩個點，但是步數不會超過較長的那條邊，想想是為啥？因為要達到目標點，橫縱方向都要走完差值，走對角線可以同時走橫縱方向，但要走的步數也絕不會小於較大的差值，不然的話在該方向沒法走完之間的距離，同時也不應該大於較大的差值，否則走的就不是最優解的話。所以兩個點之間的步數就是其橫縱坐標各自的差的絕對值中的較大一個，兩兩之間計算一下，累加起來即為所求，參見代碼如下：

class Solution {
public:
    int minTimeToVisitAllPoints(vector<vector<int>>& points) {
        int res = 0, n = points.size();
        for (int i = 0; i < n - 1; ++i) {
            res += max(abs(points[i][0] - points[i + 1][0]), abs(points[i][1] - points[i + 1][1]));
        }
        return res;
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1266

參考資料：

https://leetcode.com/problems/minimum-time-visiting-all-points/

https://leetcode.com/problems/minimum-time-visiting-all-points/discuss/436250/JavaPython-3-6-liner-and-1-liner-w-brief-explanation-and-analysis.

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