题目原文:

Decimal dominants. Given an array with n keys, design an algorithm to find all values that occur more than  n/10 times. The expected running time of your algorithm should be linear.

分析:

直观上将n个元素遍历一遍,并记录每个元素出现的次数就可以实现,虽然时间复杂度是O(n),但是空间复杂度却高达n,这肯定不是该题目的初衷。对于n个元素来说,出现n/10次的元素最多有10个,那么出现超过n/10次的元素最多不超过9个,所以需要9个额外空间auxs就能满足需求。

这9个辅助空间aux怎么使用呢?可采用俄罗斯方块的消去一行的思路。只不过这里消去一行的情况是该行中元素各不相同。

1. 遍历数组array中的每个元素array[i]

2. 如果array[i]在aux中存在,将其在aux中的计数+1

3. 如果array[i]在aux中不存在

  3.1 如果aux未满,将其放入aux中,并记录其个数为1

  3.2 如果aux已满,将aux中已经存在的各个元素的计数都减去1,直到某个元素的个数变成0,将array[i]放入aux中该位置处,并记录其个数为1

4. 出现次数超过n/10的元素在array遍历完了之后,还会继续存在于aux中,当然aux中可存在着位于array后方但出现次数不满足要求的元素。这时只需要遍历aux的同时再遍历一遍array,记录aux中各个元素在array中出现的次数,将其中出现次数真正超过n/10的元素找出来即可。

 package week3;

 import java.util.ArrayList;
import java.util.Arrays;
import edu.princeton.cs.algs4.StdRandom; public class ElemsMoreThanNDivTenTimes { private class Element{//辅助空间元素定义,用来记录元素值及其出现次数
public int element;
public int count;
public Element(int e,int c){
this.element = e;
this.count = c;
}
};
private Element[] elems = new Element[9]; //申请9个辅助空间 public ArrayList<Integer> findElements(int[] arrays){
int n = arrays.length;
for(int k=0;k<9;k++){
elems[k] = new Element(0,0); //辅助空间初始化
}
for(int i=0;i<n;i++){
int index = findIndex(arrays[i]);
if(index >= 0)
elems[index].count ++;
else
addToElems(arrays[i]);
}
return verifyElems(arrays);
} private int findIndex(int e){
for(int k = 0; k<9;k++){
if(elems[k].element == e)
return k;
else if(elems[k].count == 0){
elems[k].element = e;
return k;
}
}
return -1;
}
private void addToElems(int e){
boolean insertFlag = false;
while(!insertFlag){
for(int k=0; k<9;k++){
elems[k].count --;
if(elems[k].count <= 0){
elems[k].element = e;
elems[k].count = 1;
insertFlag = true;
break;
}
}
}
}
private ArrayList<Integer> verifyElems(int[] arrays){
int n = arrays.length;
for(int k = 0; k< 9; k++){
elems[k].count = 0;
for(int i = 0; i< n;i++){
if(arrays[i]==elems[k].element)
elems[k].count++;
}
}
ArrayList<Integer> elemList = new ArrayList<Integer>();
for(int k = 0; k< 9; k++){
if(elems[k].count > n/10)
elemList.add(elems[k].element);
}
return elemList;
} public static void main(String[] args){
int n = 20;
int[] array = new int[n];
for(int i=0;i<n;i++){
array[i] = StdRandom.uniform(n);
}
System.out.println(Arrays.toString(array));
ElemsMoreThanNDivTenTimes elems = new ElemsMoreThanNDivTenTimes();
ArrayList<Integer> elemList = elems.findElements(array);
System.out.println(elemList.toString());
}
}

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