????? java JDK里面容器類的排序算法使用的主要是插入排序和歸并排序，可能不同版本的實現有所不同，關鍵代碼如下：

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/*** Performs a sort on the section of the array between the given indices* using a mergesort with exponential search algorithm (in which the merge* is performed by exponential search). n*log(n) performance is guaranteed* and in the average case it will be faster then any mergesort in which the* merge is performed by linear search.* * @param in -* the array for sorting.* @param out -* the result, sorted array.* @param start* the start index* @param end* the end index + 1*/@SuppressWarnings("unchecked")private static void mergeSort(Object[] in, Object[] out, int start,int end) {int len = end - start;// use insertion sort for small arraysif (len <= SIMPLE_LENGTH) {for (int i = start + 1; i < end; i++) {Comparable<Object> current = (Comparable<Object>) out[i];Object prev = out[i - 1];if (current.compareTo(prev) < 0) {int j = i;do {out[j--] = prev;} while (j > start&& current.compareTo(prev = out[j - 1]) < 0);out[j] = current;}}return;}int med = (end + start) >>> 1;mergeSort(out, in, start, med);mergeSort(out, in, med, end);// merging// if arrays are already sorted - no mergeif (((Comparable<Object>) in[med - 1]).compareTo(in[med]) <= 0) {System.arraycopy(in, start, out, start, len);return;}int r = med, i = start;// use merging with exponential searchdo {Comparable<Object> fromVal = (Comparable<Object>) in[start];Comparable<Object> rVal = (Comparable<Object>) in[r];if (fromVal.compareTo(rVal) <= 0) {int l_1 = find(in, rVal, -1, start + 1, med - 1);int toCopy = l_1 - start + 1;System.arraycopy(in, start, out, i, toCopy);i += toCopy;out[i++] = rVal;r++;start = l_1 + 1;} else {int r_1 = find(in, fromVal, 0, r + 1, end - 1);int toCopy = r_1 - r + 1;System.arraycopy(in, r, out, i, toCopy);i += toCopy;out[i++] = fromVal;start++;r = r_1 + 1;}} while ((end - r) > 0 && (med - start) > 0);// copy rest of arrayif ((end - r) <= 0) {System.arraycopy(in, start, out, i, med - start);} else {System.arraycopy(in, r, out, i, end - r);}}

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??????? 看到編程珠璣上有一個很有趣的排序算法-位圖法其思想是用1位來表示[0~n-1]中的整數是否存在。1表示存在，0表示不存在。即將正整數映射到bit集合中，每一個bit代表其映射的正整數是否存在。

? 比如{1，2，3，5，8，13}使用下列集合表示：

? 0 1 1 1 0 1 0 0 1 0 0 0 0 1 0 0 0 0 0 0

偽代碼如下：

for (i in [0~n-1]) bit[i] = 0; for(i in [0~n-1]) if (i in input file) bit[i] = 1for(i in [0~n-1])if(bit[i] == 1) output i

?????? 用java 代碼嘗試下，效率果然不錯：

public class javaUniqueSort {public static int[] temp = new int[1000001];public static List<Integer> tempList = new ArrayList<Integer>();public static int count;public static void main(final String[] args) {List<Integer> firstNum = new ArrayList<Integer>();List<Integer> secondNum = new ArrayList<Integer>();for (int i = 1; i <= 1000000; i++) {firstNum.add(i);secondNum.add(i);}Collections.shuffle(firstNum);Collections.shuffle(secondNum);getStartTime();Collections.sort(firstNum);getEndTime("java sort run time ");getStartTime();secondNum = uniqueSort(secondNum);getEndTime("uniqueSort run time ");}public static List<Integer> uniqueSort(final List<Integer> uniqueList) {javaUniqueSort.tempList.clear();for (int i = 0; i < javaUniqueSort.temp.length; i++) {javaUniqueSort.temp[i] = 0;}for (int i = 0; i < uniqueList.size(); i++) {javaUniqueSort.temp[uniqueList.get(i)] = 1;}for (int i = 0; i < javaUniqueSort.temp.length; i++) {if (javaUniqueSort.temp[i] == 1) {javaUniqueSort.tempList.add(i);}}return javaUniqueSort.tempList;}public static void getStartTime() {javaShuffle.start = System.nanoTime();}public static void getEndTime(final String s) {javaShuffle.end = System.nanoTime();System.out.println(s + ": " + (javaShuffle.end - javaShuffle.start) + "ns");} }運行時間：java sort run time : 1257737334ns uniqueSort run time : 170228290ns java sort run time : 1202749828ns uniqueSort run time : 169327770ns

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如果有重復數據，可以修改下：

public class javaDuplicateSort {public static List<Integer> tempList = new ArrayList<Integer>();public static int count;public static void main(final String[] args) {Random random = new Random();List<Integer> firstNum = new ArrayList<Integer>();List<Integer> secondNum = new ArrayList<Integer>();for (int i = 1; i <= 100000; i++) {firstNum.add(i);secondNum.add(i);firstNum.add(random.nextInt(i + 1));secondNum.add(random.nextInt(i + 1));}Collections.shuffle(firstNum);Collections.shuffle(secondNum);getStartTime();Collections.sort(firstNum);getEndTime("java sort run time ");getStartTime();secondNum = uniqueSort(secondNum);getEndTime("uniqueSort run time ");}public static List<Integer> uniqueSort(final List<Integer> uniqueList) {javaDuplicateSort.tempList.clear();int[] temp = new int[200002];for (int i = 0; i < temp.length; i++) {temp[i] = 0;}for (int i = 0; i < uniqueList.size(); i++) {temp[uniqueList.get(i)]++;}for (int i = 0; i < temp.length; i++) {for (int j = temp[i]; j > 0; j--) {javaDuplicateSort.tempList.add(i);}}return javaDuplicateSort.tempList;}public static void getStartTime() {javaShuffle.start = System.nanoTime();}public static void getEndTime(final String s) {javaShuffle.end = System.nanoTime();System.out.println(s + ": " + (javaShuffle.end - javaShuffle.start) + "ns");} }

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????? 這種算法還是有很明顯的局限性的，比如說要知道數據中最大的數值，更重要的是數據的疏密程度，比如說最大值為1000000而要數組大小只有100，那么效率會下降的非常明顯。。。。。但是，使用位圖法進行排序，確實讓人眼前一亮。位圖法通常是用來存儲數據，判斷某個數據存不存在或者判斷數組是否存在重復 。

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轉載于:https://www.cnblogs.com/Kingle/archive/2012/03/09/2994679.html

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