LeetCode:1266. Minimum Time Visiting All Points

題目:
On a plane there are n points with integer coordinates points[i] = [xi, yi]. Your task is to find the minimum time in seconds to visit all points.
You can move according to the next rules:
* In one second always you can either move vertically, horizontally by one unit or diagonally (it means to move one unit vertically and one unit horizontally in one second).
* You have to visit the points in the same order as they appear in the array.

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

依照範例給的解釋,可以歸納是找出最短的移動路徑(可以斜著走)
但不管是斜著走還是直著走,都是一次移動

在switch中,如果下一點到x、到y的距離都是d,就等於直接加上d
其它的情況就是加上最遠距離,記得前面要加上方向的判斷

func minTimeToVisitAllPoints(points [][]int) int {
    var xDiff, yDiff int
    seconds := 0
    for i:=0; i < (len(points)-1);i++ {
        if xDiff = points[i+1][0] - points[i][0]; xDiff < 0{
            xDiff = -xDiff
        }      
        if yDiff = points[i+1][1] - points[i][1]; yDiff < 0{
            yDiff = -yDiff
        }      
        switch{
        case xDiff == yDiff:
            seconds += xDiff
        case xDiff > yDiff:
            seconds += xDiff
        case xDiff < yDiff:
            seconds += yDiff
        }
    }
    return seconds
}

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