用Java删除数组中所有相同的数,使之仅剩下一个,然后输出剩下的数据？ Java去除数组中的重复的数据后输出数组并把重复的数输出？
\u7f16\u5199\u7a0b\u5e8f\u5728\u88ab\u8c03\u51fd\u6570\u4e2d\u5220\u53bb\u4e00\u7ef4\u6570\u7ec4\u4e2d\u6240\u6709\u76f8\u540c\u7684\u6570\uff0c\u4f7f\u4e4b\u53ea\u5269\u4e00\u4e2a\uff0c\u88ab\u8c03\u51fd\u6570\u8fd4\u56de\u5220\u9664\u540e\u6570\u7ec4\u4e2d\u7684\u6570\u636e\u7684\u4e2a\u6570int main()
{
int a[20]={4,4,4,6,7,7,7,9,9,9,12,12,24,34,34,34,56,67,78,90};
int i=0,j=0,k=0,b[20];//b[]\u4e3a\u4fdd\u5b58\u4e0d\u76f8\u540c\u7684\u6570\u7ec4\u6210\u5458
for(i=0;i<20;i++)//\u904d\u5386\u6bcf\u4e2a\u6570\u7ec4\u6210\u5458
{
for(j=0;j<k;j++)//\u6570\u7ec4\u6210\u5458\u4e0e\u4e0d\u540c\u6570\u7ec4\u6210\u5458\u6bd4\u8f83
{
if(b[j]==a[i])//\u5982\u679c\u4e0e\u4e0d\u540c\u6570\u7ec4\u6bd4\u8f83\u6709\u76f8\u540c\u8005\uff0c\u8df3\u51fa\uff0c\u8fdb\u884c\u4e0b\u4e00\u4e2a\u6570\u7ec4\u6210\u5458\u6bd4\u8f83
break;
}
if(j==k)//\u5982\u679c\u4e0e\u4e0d\u540c\u6570\u7ec4\u7684\u6210\u5458\u90fd\u4e0d\u540c\uff0c\u5219\u52a0\u5165\u4e0d\u540c\u6570\u7ec4\uff1b
{
b[k]=a[i];
k++;
}
}
for(i=0;i<k;i++)
printf("%d,",b[i]);
getchar();
return 0;
}
\u5c31\u662f\u518d\u63a5\u4e00\u4e2a\u65b0\u7684list
\u4f60if{}\u540e\u9762\u52a0\u4e86else\u628a\u91cd\u590d\u6570\u636e\u6dfb\u52a0\u5230\u65b0list
\u7136\u540e\u904d\u5386\u5c31\u53ef\u4ee5\u4e86
/**
* 方法1 就是申请一个和nums大小一样的数组tmpNums，
* 然后遍历nums中的元素，对每个元素判断是否在tmpNums出现过，
* 如果出现过，那么就不放到新数组里面，也就是不做处理；
* 如果没有出现过，那么就把它放到新的数组里面
* 这种方法数组原来的相对顺序可以保证
* 时间复杂度是n^2和空间复杂度是n
*
*
@param nums 输入需要去重的数组
*
@return 返回去重后数组的长度
*/
public static int unique1(int[] nums) {
if (nums.length == 0) {
return 0;
}
int[] tmpNums = new int[nums.length];
tmpNums[
0] = nums[0];
int index = 1;
for (int i = 1, len = nums.length; i < len; i++) {
int j = 0;
for (; j < index; j++) {
if (tmpNums[j] == nums[i]) {
break;
}
}
if (j == index) {
tmpNums[index
++] = nums[i];
}
}
nums
= tmpNums;
return index;
}
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