LeetCode - 939. Minimum Area Rectangle (C++)
阿新 • • 發佈:2018-11-19
LeetCode - 939. Minimum Area Rectangle
Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes.
If there isn't any rectangle, return 0.
Example 1:
Input: [[1,1],[1,3],[3,1],[3,3],[2,2]] Output:4
Example 2:
Input: [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]] Output: 2
Note:
1 <= points.length <= 500
0 <= points[i][0] <= 40000
0 <= points[i][1] <= 40000
- All points are distinct.
解:
這道題目一開始的想法是,畫很多條與座標軸平行的線(比如 x = 5 出現過兩次,那麼 x = 5 這個位置就有一根線
所以改用最直接的方法,就是將所有點,也就是所有 (x, y) 的組合存起來,之後,每兩個點 (x1, y1), (x2, y2) 查詢一次有沒有 (x1, y2), (x2, y1) 兩個點存在,這樣就能構成一個矩形,計算面積,迴圈得到面積的最小值。將組合存起來最直接的想法就是 unordered_set<pair<int, int>>,但是這樣 C++ 是不知道如何計算 pair 的 hash 值的,所以需要過載 () 操作符,比較麻煩,改用效率低一些的 set 即可,AC程式碼如下:
int minAreaRect(vector<vector<int>>& points) {
bool found = false;
int psize = points.size(), min_area = INT_MAX;
set<pair<int, int>> setp; // unordered_set會報錯 "The C++ Standard doesn't provide a hash for this type."
for (auto p : points)
setp.insert(pair<int, int>(p[0], p[1]));
for (int i = 0; i < psize; i++)
for (int j = i + 1; j < psize; j++)
{
int x1 = points[i][0], y1 = points[i][1], x2 = points[j][0], y2 = points[j][1];
if (x1 == x2 || y1 == y2) continue;
if (abs(x1 - x2) * abs(y1 - y2) >= min_area) continue; // 一定不是最小矩形
if (setp.find(pair<int, int>(x1, y2)) != setp.end()
&& setp.find(pair<int, int>(x2, y1)) != setp.end()) // 另兩個點存在
{
found = true;
min_area = min(min_area, abs(x1 - x2) * abs(y1 - y2));
}
}
if (found == false) return 0;
return min_area;
}