1266. Minimum Time Visiting All Points

Easy

題目描述

Description

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

解答

Solution

/**
 * @param {number[][]} points
 * @return {number}
 */
var minTimeToVisitAllPoints = function (points) {
  let result = 0;

  for (let i = 0; i < points.length - 1; i++) {
    const currentX = points[i][0];
    const currentY = points[i][1];
    const targetX = points[i + 1][0];
    const targetY = points[i + 1][1];
    const distanceX = Math.abs(currentX - targetX);
    const distanceY = Math.abs(currentY - targetY);

    result += Math.max(distanceX, distanceY);
  }

  return result;
};

解題思路

根據題目的描述，移動方式只有

1. 水平移動
2. 垂直移動
3. 斜走

而斜走可以一次移動 X 和 Y，故以斜走為優先，當不行斜走時候才水平或垂直移動。

關鍵在此處

const distanceX = Math.abs(currentX - targetX);
const distanceY = Math.abs(currentY - targetY);

result += Math.max(distanceX, distanceY); // 比較 X 和 Y 各自的移動哪個會花比較多步，先斜走走完共通部分再走完剩下的部分

心得

題目看起來有點拗，但把 目前 X 座標、目前 Y 座標、目標 X 座標、目標 Y 座標 各自先定義好後，語意就清晰多了!