QuickHull C++ solution 29ms

  • 1

    Pseudo code (from Wikipedia):
    Input = a set S of n points
    Assume that there are at least 2 points in the input set S of points
    QuickHull (S)
    // Find convex hull from the set S of n points
    Convex Hull := {}
    Find left and right most points, say A & B, and add A & B to convex hull
    Segment AB divides the remaining (n-2) points into 2 groups S1 and S2
    where S1 are points in S that are on the right side of the oriented line from A to B,
    and S2 are points in S that are on the right side of the oriented line from B to A
    FindHull (S1, A, B)
    FindHull (S2, B, A)
    FindHull (Sk, P, Q)
    // Find points on convex hull from the set Sk of points
    // that are on the right side of the oriented line from P to Q
    If Sk has no point, then return.
    From the given set of points in Sk, find farthest point, say C, from segment PQ
    Add point C to convex hull at the location between P and Q
    Three points P, Q, and C partition the remaining points of Sk into 3 subsets: S0, S1, and S2
    where S0 are points inside triangle PCQ, S1 are points on the right side of the oriented
    line from P to C, and S2 are points on the right side of the oriented line from C to Q.
    FindHull(S1, P, C)
    FindHull(S2, C, Q)
    Output = convex hull

    class Solution {
        static bool mycmp(Point &a, Point &b) {
            return a.x < b.x;
        int testSide(Point &a, Point &b, Point &c) {
            // cross product of (AB and AC vectors)
            return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
        double distPointLine(Point &A, Point &B, Point &C) {
            // dist(line: ax+by+c=0, and point(x0, y0)): (a*x0 + b*y0 + c)/sqrt(a^2+b^2)
            // line: (y2-y1)*x - (x2-x1)*y + x2*y1 - y2*x1 = 0
            int a = B.y - A.y, b = B.x - A.x;
            return abs((a*C.x - b*C.y + B.x*A.y - B.y*A.x)/sqrt(a*a + b*b));
        void FindHull(vector<Point> &points, Point &A, Point &B) {
            if (points.empty())
            int idx = 0;
            double dist = distPointLine(A, B, points[0]);
            for (int i=1; i<points.size(); i++) {
                if (distPointLine(A, B, points[i]) > dist) {
                    dist = distPointLine(A, B, points[i]);
                    idx = i;
            vector<Point> R, T;
            for (int i=0; i<points.size(); i++) {
                if (i != idx) {
                    int tmp = testSide(A, points[idx], points[i]);
                    if (tmp >= 0)
                    else {
                        tmp = testSide(points[idx], B, points[i]);
                        if (tmp >= 0)
            FindHull(R, A, points[idx]);
            FindHull(T, points[idx], B);
        vector<Point> ret;
        vector<Point> outerTrees(vector<Point>& points) {
            // find the convex hull; use QuickHull algorithm
            if (points.size() <= 1)
                return points;
            // find the left most and right most two points
            sort(points.begin(), points.end(), mycmp);
            // test whether a point on the left side right side or on the line
            vector<Point> Left, Right, Online;
            for (int i=1; i<points.size()-1; i++) {
                int tmp = testSide(points[0], points.back(), points[i]);
                if (tmp < 0)
                else if (tmp > 0)
            // if Upper or Down is empty, Online should be pushed into ret
            if (Left.empty() || Right.empty())
                for (int i=0; i<Online.size(); i++)
            FindHull(Left, points[0], points.back());
            FindHull(Right, points.back(), points[0]);
            return ret;

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.