城市间最大非重叠桥梁的算法

Question

河的两边有相同数量的城市。从一侧的城市到另一侧的城市的桥梁由1#y3代表，其中下方的城市1在上侧具有通往城市3的桥梁。我们必须找到非重叠桥的最大数量。因此，对于输入1#y2, 2#y4, 3#y1, 4#y5, 5#y3, 6#y6，4将1#y2, 2#y4, 4#y5, 6#y6与public static int maxNonOverlappingBridges(String input1[]) { int result = 0; for (int i = 0; i < input1.length; i++) { int total = 1; int notCrossing = Integer.parseInt(input1[i].substring(input1[i].length() - 1)); for (int j = 0; j < input1.length; j++) { if (j < i) { if (Integer.parseInt(input1[j].substring(input1[j].length() - 1)) < notCrossing) { total += 1; notCrossing = Integer.parseInt(input1[j].substring(input1[j].length() - 1)); } } else if (j > i) { if (Integer.parseInt(input1[j].substring(input1[j].length() - 1)) > notCrossing) { total += 1; notCrossing = Integer.parseInt(input1[j].substring(input1[j].length() - 1)); } } else { notCrossing = Integer.parseInt(input1[j].substring(input1[j].length() - 1)); } } if (total > result) result = total; } return result; } 不重叠。

这是我的代码 -

SELECT emp_name, position, absent, 
(select sum(absent) from employee) as sum, 
(absent*100)/(select sum(absent) from employee) AS 'percentage' 
FROM employee

是否有更优化的算法？

Answer 1

这在 O（nLog（n））中运行。

return {"project":o}

希望这有帮助。

Answer 2

O（nLog（n）） ...

的简单解决方案

openjdk\hotspot\src\share\vm\runtime\jniHandles.cpp

Answer 3

public static int bridge_count(String[] input1,int input2)
{
    int[][] track=new int[input2][input2];
    int count=0;

    for(int i=0;i<input1.length;i++)
    {
        String cityCon=input1[i];
        String cityArray[]=cityCon.split("#");
        int fst=Integer.parseInt(cityArray[0]);
        int sec=Integer.parseInt(cityArray[1]);
        track[fst-1][sec-1]=-1;
    }

    int found=0;
    int maxCol=-1;

    for(int c=0;c<input2;c++)
    {
        found=-1;
        for(int j=0;j<input2;j++)
        {
            if(track[c][j]==-1)
            {
             found=j;
             break;
            }
        }    

        if(found!=-1 && maxCol<=found)
        {
            count++;
            maxCol=found;
        }    
    }
    return count;
}