356. Line Reflection

来源：互联网发布：淘宝骑行头盔编辑：程序博客网时间：2024/05/16 11:45

Given n points on a 2D plane, find if there is such a line parallel to y-axis that reflect the given points.

Example 1:

Given points = [[1,1],[-1,1]], return true.

Example 2:

Given points = [[1,1],[-1,-1]], return false.

Follow up:
Could you do better than O(n2)?

Hint:

Find the smallest and largest x-value for all points.
If there is a line then it should be at y = (minX + maxX) / 2.
For each point, make sure that it has a reflected point in the opposite side.

Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.

解题思路在hint中已经给出，关键是怎么求opposit point，和怎么存point了。开始是新建一个point类，用hashmap<integer, list<point>>存point。求opposit point的时候先算中点，再用中点求opposit point，这样就得用double计算。以上两种情况是最初思路，代码如下：

public class Solution {    private class Point {        int x;        int y;        public Point(int px, int py) {            this.x = px;            this.y = py;        }    }        public boolean isReflected(int[][] points) {        if (points == null || points.length == 0 || points[0].length == 0) {            return true;        }        double minX = Integer.MAX_VALUE, maxX = Integer.MIN_VALUE;        HashMap<Integer, List<Point>> map = new HashMap<Integer, List<Point>>();        for (int[] point: points) {            if (point[0] < minX) {                minX = (double)point[0];            }            if (point[0] > maxX) {                maxX = (double)point[0];            }            if (!map.containsKey(point[0]))                map.put(point[0], new ArrayList<Point>());            map.get(point[0]).add(new Point(point[0], point[1]));        }        double mid = minX + (maxX - minX) / 2;        for (int[] point: points) {            double reflect = mid + mid - point[0];            boolean find = false;            if (map.containsKey((int)reflect)) {                for (Point p: map.get((int)reflect)) {                    if (point[1] == p.y) {                        find = true;                    }                }                if (find == false) {                    return false;                }            } else {                return false;            }        }        return true;    }}

有两点可以改进：第一，改用HashSet.add(Arrays.hashCode(point)) 存点；第二，求对称点时，用minX + maxX求出和，因为是对称点，中点都是一样的，那么sum也肯定是一样的，sum减其中一个点的x，就是另一个点的x。代码如下：

public class Solution {    public boolean isReflected(int[][] points) {        HashSet<Integer> pointSet = new HashSet<>();        int sum;        int maxX, minX;                minX = Integer.MAX_VALUE;        maxX = Integer.MIN_VALUE;        for(int[] point:points) {            maxX = Math.max(maxX, point[ 0 ]);            minX = Math.min(minX, point[ 0 ]);            pointSet.add(Arrays.hashCode(point));        }                sum = maxX+minX;        for(int[] point:points) {            if(!pointSet.contains(Arrays.hashCode(new int[]{sum-point[ 0 ], point[ 1 ]}))) {                return false;            }        }        return true;    }}

0 0