最近发现一个有意思的插值器PathInterpolator，传一个或两个点进去就可当插值器使用。可以看一下两个构造函数，生成的path是一个贝赛尔曲线。
/** * Create an interpolator for a quadratic Bezier curve. The end points * <code>(0, 0)</code> and <code>(1, 1)</code> are assumed. * * @param controlX The x coordinate of the quadratic Bezier control point. * @param controlY The y coordinate of the quadratic Bezier control point. */ public PathInterpolator(float controlX, float controlY) { initQuad(controlX, controlY); } /** * Create an interpolator for a cubic Bezier curve. The end points * <code>(0, 0)</code> and <code>(1, 1)</code> are assumed. * * @param controlX1 The x coordinate of the first control point of the cubic Bezier. * @param controlY1 The y coordinate of the first control point of the cubic Bezier. * @param controlX2 The x coordinate of the second control point of the cubic Bezier. * @param controlY2 The y coordinate of the second control point of the cubic Bezier. */ public PathInterpolator(float controlX1, float controlY1, float controlX2, float controlY2) { initCubic(controlX1, controlY1, controlX2, controlY2); } private void initQuad(float controlX, float controlY) { Path path = new Path(); path.moveTo(0, 0); path.quadTo(controlX, controlY, 1f, 1f); initPath(path); } private void initCubic(float x1, float y1, float x2, float y2) { Path path = new Path(); path.moveTo(0, 0); path.cubicTo(x1, y1, x2, y2, 1f, 1f); initPath(path); }两个构造函数生成的贝塞尔曲线都是从，(0,0)到(1,1)点的Path。
曲线生成好了之后，会将其重新解析成若干个点的数组
private void initPath(Path path) { float[] pointComponents = path.approximate(PRECISION); int numPoints = pointComponents.length / 3; if (pointComponents[1] != 0 || pointComponents[2] != 0 || pointComponents[pointComponents.length - 2] != 1 || pointComponents[pointComponents.length - 1] != 1) { //这个Path的起始点必须是(0,0),终止点必须是(1,1) throw new IllegalArgumentException("The Path must start at (0,0) and end at (1,1)"); } mX = new float[numPoints]; mY = new float[numPoints]; float prevX = 0; float prevFraction = 0; int componentIndex = 0; for (int i = 0; i < numPoints; i++) { float fraction = pointComponents[componentIndex++]; float x = pointComponents[componentIndex++]; float y = pointComponents[componentIndex++]; if (fraction == prevFraction && x != prevX) { //判断这个Path是否连续，这个可以借鉴一下 throw new IllegalArgumentException( "The Path cannot have discontinuity in the X axis."); } if (x < prevX) { throw new IllegalArgumentException("The Path cannot loop back on itself."); } mX[i] = x; mY[i] = y; prevX = x; prevFraction = fraction; } }作为插值器，自然需要复写getInterpolation
@Override public float getInterpolation(float t) { if (t <= 0) { return 0; } else if (t >= 1) { return 1; } // Do a binary search for the correct x to interpolate between. int startIndex = 0; int endIndex = mX.length - 1; while (endIndex - startIndex > 1) { int midIndex = (startIndex + endIndex) / 2; if (t < mX[midIndex]) { endIndex = midIndex; } else { startIndex = midIndex; } } float xRange = mX[endIndex] - mX[startIndex]; if (xRange == 0) { return mY[startIndex]; } float tInRange = t - mX[startIndex]; float fraction = tInRange / xRange; float startY = mY[startIndex]; float endY = mY[endIndex]; return startY + (fraction * (endY - startY)); }算法里面用二分法找到了对应的index，返回值是一个线性的计算，很简单。
最后看一下path.approximate的算法。
// Returns a float[] with each point along the path represented by 3 floats // * fractional length along the path that the point resides // * x coordinate // * y coordinate // Note that more than one point may have the same length along the path in // the case of a move. // NULL can be returned if the Path is empty. static jfloatArray approximate(JNIEnv* env, jclass, jlong pathHandle, float acceptableError) { SkPath* path = reinterpret_cast<SkPath*>(pathHandle); SkASSERT(path); SkPath::Iter pathIter(*path, false); SkPath::Verb verb; SkPoint points[4]; std::vector<SkPoint> segmentPoints; std::vector<float> lengths; float errorSquared = acceptableError * acceptableError; while ((verb = pathIter.next(points, false)) != SkPath::kDone_Verb) { createVerbSegments(verb, points, segmentPoints, lengths, errorSquared); } if (segmentPoints.empty()) { int numVerbs = path->countVerbs(); if (numVerbs == 1) { addMove(segmentPoints, lengths, path->getPoint(0)); } else { // Invalid or empty path. Fall back to point(0,0) addMove(segmentPoints, lengths, SkPoint()); } } float totalLength = lengths.back(); if (totalLength == 0) { // Lone Move instructions should still be able to animate at the same value. segmentPoints.push_back(segmentPoints.back()); lengths.push_back(1); totalLength = 1; } size_t numPoints = segmentPoints.size(); size_t approximationArraySize = numPoints * 3; float* approximation = new float[approximationArraySize]; int approximationIndex = 0; for (size_t i = 0; i < numPoints; i++) { const SkPoint& point = segmentPoints[i]; approximation[approximationIndex++] = lengths[i] / totalLength; approximation[approximationIndex++] = point.x(); approximation[approximationIndex++] = point.y(); } jfloatArray result = env->NewFloatArray(approximationArraySize); env->SetFloatArrayRegion(result, 0, approximationArraySize, approximation); delete[] approximation; return result; } };太深入我没有继续研究，不过返回的点数组是这样的
for (size_t i = 0; i < numPoints; i++) { const SkPoint& point = segmentPoints[i]; approximation[approximationIndex++] = lengths[i] / totalLength; approximation[approximationIndex++] = point.x(); approximation[approximationIndex++] = point.y(); }