Leetcode1266. Minimum Time Visiting All Points

标签：row   second   inf   seconds   最小   ota   return   定义   cond   

public int minTimeToVisitAllPoints(int[][] points) {
    int rowLength = points.length; //获取行数
    int count = 0;
    for (int i = 0; i < rowLength - 1; i++) {
        int min = -1; //定义横坐标与纵坐标之间最小值
        int rowdif = Math.abs(points[i][0] - points[i + 1][0]);//行坐标值之间差值
        int coldif = Math.abs(points[i][1] - points[i + 1][1]);//列左边之间差值
        if (rowdif > coldif) { //行列之间差值取较小值是走斜步的步数
            min = coldif;
        } else {
            min = rowdif;
        }
        count = count + min + Math.abs(rowdif - coldif);
    }
    //    System.out.println("Total time="+count+"seconds");
    return count;
}
我的解法速度不是很快

好吧 我的两个值相加实际上就是两对差值里的最大值

Leetcode1266. Minimum Time Visiting All Points

标签：row   second   inf   seconds   最小   ota   return   定义   cond   

原文地址：https://www.cnblogs.com/chengxian/p/12204022.html