去掉一个已经排好序的数组的重复数字呢

去掉一个已经排好序的数组的重复数字,速度尽量快

import il.*;

public class DelCopy{

public static void main(String[] args)

{

int[] oNums ={1,2,2,3,3,3,4,4,4,4,5,5,5,5,5};

List Integer nNums = new ArrayList Integer();

int temp =oNums[0];

d(temp);

for(int i=1;i ngth;i++)

{

if(oNums[i]temp)

{

d(oNums[i]);

temp=oNums[i];

}

}

int(nNums);

}

}

Java code

public class Test {

public static List printMap1(String[] s) {

ListString list = new LinkedListString ();

ListString list2 = List(s);

dAll(list2);

return list;

}

//去除数组中重复数据

public static String[] array_unique(String[] a) {

// array_unique

ListString list = new LinkedListString ();

for (int i = 0; i ngth; i++) {

if (!ntains(a[i])) {

d(a[i]);

}

}

return (String[]) Array(new String[ze()]);

}

public static void main(String[] args) {

String[] ar = { \"cc\", \"dd\", \"cc\", \"ee\", \"ff\", \"gg\", \"cc\", \"ee\"};

String[] s = array_unique(ar);

for (String aa : s) {

intln(aa);

}

}

}

int[] nums1 ={1,2,2,3,3,3,4,4,4,4,5,5,5,5,5};

final int LENGTH = ngth;

int[] nums2 = new int[LENGTH];

nums2[0] = nums1[0];

if(nums1[0]==nums1[LENGTH-1]){

intln(nums2);

}else{

int j = 1;

for(int i=1;i LENGTH;i++){

if(nums1[i]==nums1[i-1]){

continue;

}else{

nums2[j] = nums1[i];

j++;

}

}

}

Java code

而《纽约邮报》的专栏作家皮特-维克希对于这篇报道的真实性产生了深深的质疑 public class Sample

{

public static void main(String[] args)

{

int[] array = {1, 2, 3, 3, 5, 6, 7, 8, 8, 8, 10, 12, 13, 13, 15, 20};

long s = noTime();

int len = ngth;

int[] temp = new int[len];

int i = 0;

int j = 0;

temp[j] = array[i];

while (i len)

{

if (array[i++] != temp[j])

{

temp[++j] = array[i - 1];

}

}

int[] result = new int[++j];

raycopy(temp, 0, result, 0, j);

long e = noTime();

intln(\"TIME=\" + (e - s));

// for (int k = 0; k ngth; k++)

// {

// intln(result[k]);

// }

}

}