/**
*
* @author gentleKay
* Given an array of integers, every element appears three times except for one. Find that single one.
* Note:
* Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
*
* 给定一个整数数组,除一个元素外,每个元素都出现三次。找到那个。
* 注:
* 您的算法应该具有线性运行时复杂性。你能在不使用额外内存的情况下实现它吗?
*/
获取map集合中键和值的三种方式:
https://www.cnblogs.com/strive-19970713/p/11282676.html
import java.util.*; /** * * @author gentleKay * Given an array of integers, every element appears three times except for one. Find that single one. * Note: * Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? * * 给定一个整数数组,除一个元素外,每个元素都出现三次。找到那个。 * 注: * 您的算法应该具有线性运行时复杂性。你能在不使用额外内存的情况下实现它吗? */ public class Main27 { public static void main(String[] args) { // TODO Auto-generated method stub int[] A = {3,3,3, 5,5,5, 6,6,6,7,7,8,7}; System.out.println(Main27.singleNumber(A)); } public static int singleNumber(int[] A) { Map<Integer, Object> map = new HashMap<>(); for (int i=0;i<A.length;i++) { if (map.containsKey(A[i])) { map.put(A[i], 2); //一旦有map里面重复这个key, 就将它的值改成 2 或者其他值。 }else { map.put(A[i], 1); //没有包含这个key键的话, 就将这个键的值为1。 } } //方法一 // Set<Integer> set = map.keySet(); // Iterator<Integer> it = set.iterator(); // while (it.hasNext()) { // Integer key = (Integer)it.next(); // Integer value = (Integer)map.get(key); // if (value == 1) { // return key; // } // } //方法二
// Set<Integer> set = map.keySet(); // for (Integer i : set) { // Integer value = (Integer) map.get(i); // if (value == 1) { // return i; // } // } //方法三 Set<Map.Entry<Integer, Object>> set = map.entrySet(); for (Map.Entry<Integer, Object> m : set) { if ((Integer)m.getValue() == 1) { return m.getKey(); } } return 0; } }