LeetCode 939. Minimum Area Rectangle (最小面积矩形)

题目标签：HashMap

　　题目给了我们一组 xy 上的点坐标，让我们找出能组成矩形里最小面积的那个。

　　首先遍历所有的点，把x 坐标当作key 存入map，把重复的y坐标组成set，当作value 存入map。

　　然后遍历所有的点，找出对角的两个点，再去map里确认是否存在剩下的两个对角点，计算面积，一直保留最小的那个面积。

Java Solution:

Runtime: 214 ms, faster than 78.80%

Memory Usage: 61.5 MB, less than 5.97%

完成日期：03/27/2019

关键点：找对角2个点，方便判断和操作。

class Solution {
    public int minAreaRect(int[][] points) {
        Map<Integer, Set<Integer>> map = new HashMap<>();
        int minArea = Integer.MAX_VALUE;
        
        // save each point into map by key as x, value as multiple y
        for(int[] point : points)
        {
            if(!map.containsKey(point[0]))
                map.put(point[0], new HashSet<>());
            
            map.get(point[0]).add(point[1]);
        }
        
        // find 2 diagonal points and then find the other 2 diagonal points to calculate the area
        for(int[] point1 : points)
        {
            for(int[] point2: points)
            {
                if(point1[0] == point2[0] || point1[1] == point2[1]) // if point1 and point2 are not diagonal
                    continue;
                
                if(map.get(point1[0]).contains(point2[1]) && map.get(point2[0]).contains(point1[1])) // if find the other 2 diagonal points
                    minArea = Math.min(minArea, Math.abs(point2[0] - point1[0]) * Math.abs(point2[1] - point1[1]));
            }
        }
        
        return minArea == Integer.MAX_VALUE ? 0 : minArea;
    }
}