获取数组中最大数量的索引 [英] Getting indexes of maximum number in array
问题描述
我有一个数组,其中包含排名。
I have an array containing numbers which are ranks.
这样的事情:
0 4 2 0 1 0 4 2 0 4 0 2
此处 0
对应最低排名 max
数字对应最高排名。可能有多个索引包含最高排名。
Here 0
corresponds to the lowest rank and max
number corresponds to highest rank. There may be multiple indexes containing highest rank.
我想找到数组中所有最高排名的索引。我用以下代码实现了:
I want to find index of all those highest rank in array. I have achieved with following code:
import java.util.*;
class Index{
public static void main(String[] args){
int[] data = {0,4,2,0,1,0,4,2,0,4,0,2};
int max = Arrays.stream(data).max().getAsInt();
ArrayList<Integer> indexes = new ArrayList<Integer>();
for(int i=0;i<12;i++){
if(data[i]==max){
indexes.add(i);
}
}
for(int j=0;j<indexes.size();j++){
System.out.print(indexes.get(j)+" ");
}
System.out.println();
}
}
我的结果为: 1 6 9
有没有比这更好的方法?
Is there any better way than this ?
因为,在我的情况下,可能有一个包含数百万个元素的数组,因此我对性能有一些问题。
Because, In my case there may be an array containing millions of elements due to which I have some issue regarding performance.
所以,
任何建议都表示赞赏。
推荐答案
一种方法是简单地沿阵列进行单次传递并跟踪最高数量的所有索引。如果当前条目小于目前为止看到的最高数字,那么no-op。如果当前条目与看到的最大数字相同,则添加该索引。否则,我们已经看到了一个新的最高数字,我们应该扔掉我们最旧的最高数字列表并开始一个新的数字。
One approach would be to simply make a single pass along the array and keep track of all indices of the highest number. If the current entry be less than the highest number seen so far, then no-op. If the current entry be the same as the highest number seen, then add that index. Otherwise, we have seen a new highest number and we should throw out our old list of highest numbers and start a new one.
int[] data = {0,4,2,0,1,0,4,2,0,4,0,2};
int max = Integer.MIN_VALUE;
List<Integer> vals = new ArrayList<>();
for (int i=0; i < data.length; ++i) {
if (data[i] == max) {
vals.add(i);
}
else if (data[i] > max) {
vals.clear();
vals.add(i);
max = data[i];
}
}
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