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)?

Solution1：selfencode + Hashset

思路： first pass先求出中心線并將每個點存入hashset，second pass 看是否每個點在中心線的對面對應(yīng)位置存在另一個點。
關(guān)于點存入hashset的方式：這里自己編碼了。因為hashset的key必須是id型，不能直接存collection，但collection object可以通過hashCode轉(zhuǎn)成id存入，如solution2.
Time Complexity: O(N) Space Complexity: O(N)

Solution2：list_hashcode + Hashset

思路： 思路同1，但實現(xiàn)上直接將collection 通過 hashCode轉(zhuǎn)成id存入hashset

Solution1 Code：

class Solution {
    public boolean isReflected(int[][] points) {
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        HashSet<String> set = new HashSet<>();
        
        for(int[] p: points){
            max = Math.max(max, p[0]);
            min = Math.min(min, p[0]);
            String str = p[0] + "a" + p[1];
            set.add(str);
        }
        
        int sum = max + min;
        for(int[] p: points){
            String str = (sum - p[0]) + "a" + p[1];
            if( !set.contains(str)) {
                return false;
            }
        }
        return true;
    }
}

Solution2 Code：

class Solution2 {
    public boolean isReflected(int[][] points) {
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        HashSet<Integer> set = new HashSet<>();
        
        for(int[] p: points){
            max = Math.max(max, p[0]);
            min = Math.min(min, p[0]);
            set.add(Arrays.hashCode(p));
        }
        
        int sum = max + min;
        for(int[] p: points){
            String str = (sum - p[0]) + "a" + p[1];
            if(!set.contains(Arrays.hashCode(new int[]{sum - p[0], p[1]}))) {
                return false;
            }
        }
        return true;
    }
}