标签：move tor mini red nbsp The HERE 90度 points
Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes.
If there isn‘t any rectangle, return 0.
Example 1:
Input: [[1,2],[2,1],[1,0],[0,1]] Output: 2.00000 Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.
Input: [[0,1],[2,1],[1,1],[1,0],[2,0]] Output: 1.00000 Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.
这题也卡住了，主要是如何判断平面上4个点是矩形？
可以先判断它是个平行四边形，然后判断它的一个角是90度。
x0,x1,x2,x3
y0,y1,y2,y3
首先在不知道顺序的情况下，需要用全部的次序遍历。就是说1，2，3，4； 1，2，4，3；1，4，2，3；1，4，3，2的全排列。
其次判断直角。
#define ALL(x) (x).begin(), (x).end() #define FOR(i, a, b) for (remove_cv<remove_reference<decltype(b)>::type>::type i = (a); i < (b); i++) #define REP(i, n) FOR(i, 0, n) class Solution { public: double minAreaFreeRect(vector<vector<int>>& points) { unordered_map<int, unordered_set<int>> c; for (auto &x:points)c[x[0]].insert(x[1]); long n = points.size(), x0, y0, x1, y1,x2,y2,x3,y3, r = LONG_MAX; REP(i, n) { x0 = points[i][0]; y0 = points[i][1]; REP(j, n) { x1 = points[j][0]; y1 = points[j][1]; REP(k, n) if (k != i && k != j) { x2 = points[k][0]; y2 = points[k][1]; REP(l, n) if (l != i && l != j && l != k) { x3 = points[l][0]; y3 = points[l][1]; if (x1-x0==x2-x3 && y1-y0==y2-y3 && x3-x0==x2-x1 && y3-y0==y2-y1) if ((x1-x0)*(x3-x0)+(y1-y0)*(y3-y0)==0) r = min(r, abs((x1 - x0) * (y3 - y0) - (y1 - y0) * (x3 - x0))); } } } } return r == LONG_MAX ? 0 : r; } };
LC 963. Minimum Area Rectangle II
标签：move tor mini red nbsp The HERE 90度 points
原文地址：https://www.cnblogs.com/ethanhong/p/10165825.html
