(Easy) Valid Boomerang - LeetCode

Description:

A boomerang is a set of 3 points that are all distinct and not in a straight line.

Given a list of three points in the plane, return whether these points are a boomerang.

Example 1:

Input: [[1,1],[2,3],[3,2]]
Output: true

Example 2:

Input: [[1,1],[2,2],[3,3]]
Output: false

Note:

1. points.length == 3
2. points[i].length == 2
3. 0 <= points[i][j] <= 100

Accepted

9,468

Submissions

25,229

Solution:

class Solution {
   public boolean isBoomerang(int[][] points) {

       if(points ==null||points.length==0 ||points[0].length==0)
        return false;

       boolean line = true;
       boolean vert = true;
       for(int i =0; i<points.length-1; i++){
           if(points[i][1]!=points[i+1][1]){
               line = false;

           }

           if(points[i][0]!=points[i+1][0]){
               vert = false;
           }
       }

        List<Float> list = new ArrayList<Float>();
        float y=  0;
        float x = 0;
        float k = 0;
        List<String> dup_list = new ArrayList<String>();

        for(int i =0; i<points.length; i++){
            boolean zero_flag = false;
            boolean yuan_dian_check= false;
            String tmp ="";
            for(int j = 0; j<points[i].length; j++){
                 tmp = tmp+Integer.toString(points[i][j]);
                if(j==0){

                    if(points[i][j]==0){

                        zero_flag =true;

                    }
                    else{

                        x= points[i][j]-0;
                    }
                }

                else{

                    if(zero_flag&&points[i][j]==0){

                       yuan_dian_check = true;
                    }
                    y = points[i][j] - 0;

                }

                if(zero_flag){
                k = Integer.MAX_VALUE;
            }
            else{
                k = y/x;
            }

            }

            if(dup_list.contains(tmp)){
                return false;
            }
            else{

                dup_list.add(tmp);
            }
            if(!yuan_dian_check){
                if(!list.contains(k)){

                    list.add(k);
                }

            }
        } 

       if(list.size()<=1||vert||line){
           return false;
       }
       return true;
    }

}

原文地址：https://www.cnblogs.com/codingyangmao/p/11429087.html