path ops -- use standard SkTQSort

src/pathops/SkDQuadIntersection.cpp

 // Another approach is to start with the implicit form of one curve and solve
 // (seek implicit coefficients in QuadraticParameter.cpp
 // by substituting in the parametric form of the other.
 // The downside of this approach is that early rejects are difficult to come by.
 // http://planetmath.org/encyclopedia/GaloisTheoreticDerivationOfTheQuarticFormula.html#step
 
 
 #include "SkDQuadImplicit.h"
 #include "SkIntersections.h"
 #include "SkPathOpsLine.h"
 #include "SkQuarticRoot.h"
 #include "SkTDArray.h"
-#include "TSearch.h"
+#include "SkTSort.h"
 
 /* given the implicit form 0 = Ax^2 + Bxy + Cy^2 + Dx + Ey + F
  * and given x = at^2 + bt + c  (the parameterized form)
  *           y = dt^2 + et + f
  * then
  * 0 = A(at^2+bt+c)(at^2+bt+c)+B(at^2+bt+c)(dt^2+et+f)+C(dt^2+et+f)(dt^2+et+f)+D(at^2+bt+c)+E(dt^2+et+f)+F
  */
 
 static int findRoots(const SkDQuadImplicit& i, const SkDQuad& q2, double roots[4],
         bool oneHint, int firstCubicRoot) {
@@ skipping 138 matching lines @@
         }
     }
     int tCount = tsFound.count();
     if (tCount <= 0) {
         return true;
     }
     double tMin, tMax;
     if (tCount == 1) {
         tMin = tMax = tsFound[0];
     } else if (tCount > 1) {
-        QSort<double>(tsFound.begin(), tsFound.end() - 1);
+        SkTQSort<double>(tsFound.begin(), tsFound.end() - 1);
         tMin = tsFound[0];
         tMax = tsFound[tsFound.count() - 1];
     }
     SkDPoint end = q2.xyAtT(t2s);
     bool startInTriangle = hull.pointInHull(end);
     if (startInTriangle) {
         tMin = t2s;
     }
     end = q2.xyAtT(t2e);
     bool endInTriangle = hull.pointInHull(end);
@@ skipping 306 matching lines @@
         }
         if (lowestIndex < 0) {
             break;
         }
         insert(roots1Copy[lowestIndex], roots2Copy[closest[lowestIndex]],
                 pts1[lowestIndex]);
         closest[lowestIndex] = -1;
     } while (++used < r1Count);
     return fUsed;
 }