1/* 2 * Copyright 2014 The Android Open Source Project 3 * 4 * Licensed under the Apache License, Version 2.0 (the "License"); 5 * you may not use this file except in compliance with the License. 6 * You may obtain a copy of the License at 7 * 8 * http://www.apache.org/licenses/LICENSE-2.0 9 * 10 * Unless required by applicable law or agreed to in writing, software 11 * distributed under the License is distributed on an "AS IS" BASIS, 12 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. 13 * See the License for the specific language governing permissions and 14 * limitations under the License. 15 */ 16 17package android.support.v7.graphics; 18 19import java.util.Arrays; 20 21/** 22 * Class which provides a histogram for RGB values. 23 */ 24final class ColorHistogram { 25 26 private final int[] mColors; 27 private final int[] mColorCounts; 28 private final int mNumberColors; 29 30 /** 31 * A new {@link ColorHistogram} instance. 32 * 33 * @param pixels array of image contents 34 */ 35 ColorHistogram(final int[] pixels) { 36 // Sort the pixels to enable counting below 37 Arrays.sort(pixels); 38 39 // Count number of distinct colors 40 mNumberColors = countDistinctColors(pixels); 41 42 // Create arrays 43 mColors = new int[mNumberColors]; 44 mColorCounts = new int[mNumberColors]; 45 46 // Finally count the frequency of each color 47 countFrequencies(pixels); 48 } 49 50 /** 51 * @return number of distinct colors in the image. 52 */ 53 int getNumberOfColors() { 54 return mNumberColors; 55 } 56 57 /** 58 * @return an array containing all of the distinct colors in the image. 59 */ 60 int[] getColors() { 61 return mColors; 62 } 63 64 /** 65 * @return an array containing the frequency of a distinct colors within the image. 66 */ 67 int[] getColorCounts() { 68 return mColorCounts; 69 } 70 71 private static int countDistinctColors(final int[] pixels) { 72 if (pixels.length < 2) { 73 // If we have less than 2 pixels we can stop here 74 return pixels.length; 75 } 76 77 // If we have at least 2 pixels, we have a minimum of 1 color... 78 int colorCount = 1; 79 int currentColor = pixels[0]; 80 81 // Now iterate from the second pixel to the end, counting distinct colors 82 for (int i = 1; i < pixels.length; i++) { 83 // If we encounter a new color, increase the population 84 if (pixels[i] != currentColor) { 85 currentColor = pixels[i]; 86 colorCount++; 87 } 88 } 89 90 return colorCount; 91 } 92 93 private void countFrequencies(final int[] pixels) { 94 if (pixels.length == 0) { 95 return; 96 } 97 98 int currentColorIndex = 0; 99 int currentColor = pixels[0]; 100 101 mColors[currentColorIndex] = currentColor; 102 mColorCounts[currentColorIndex] = 1; 103 104 if (pixels.length == 1) { 105 // If we only have one pixel, we can stop here 106 return; 107 } 108 109 // Now iterate from the second pixel to the end, population distinct colors 110 for (int i = 1; i < pixels.length; i++) { 111 if (pixels[i] == currentColor) { 112 // We've hit the same color as before, increase population 113 mColorCounts[currentColorIndex]++; 114 } else { 115 // We've hit a new color, increase index 116 currentColor = pixels[i]; 117 118 currentColorIndex++; 119 mColors[currentColorIndex] = currentColor; 120 mColorCounts[currentColorIndex] = 1; 121 } 122 } 123 } 124 125} 126