[leetCode/JS] 1266. Minimum Time Visiting All Points

난이도 [ 😊 ] Easy

 

문제 설명 

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.

 

 

입출력 예

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

 

 

---

내 솔루션

/**
 * https://leetcode.com/problems/minimum-time-visiting-all-points
 * @param {number[][]} points
 * @return {number}
 */
var minTimeToVisitAllPoints = function (points) {
    let awnser = 0;
    for (let i = 0; i < points.length - 1; i++) {
        let cur = points[i];
        let next = points[i + 1];
        awnser += Math.max(
            Math.abs(next[0] - cur[0]), Math.abs(next[1] - cur[1])
        );
    }

    return awnser;
};

 

감상평

• 이지 문제가 가장 재미있다.