标签：follow dup for ++ UNC remove XA stat nbsp
Remove Duplicates from Sorted Array II
描述
Follow up for ”Remove Duplicates”: What if duplicates are allowed at most twice?
For example, Given sorted array A = [1,1,1,2,2,3],
Your function should return length = 5, and A is now [1,1,2,2,3]
分析
加一个变量记录一下元素出现的次数即可。这题因为是已经排序的数组，所以一个变量即可解
决。如果是没有排序的数组，则需要引入一个 hashmap 来记录出现次数。
代码
1 public class solution { 2 3 public static void main(String[] args) { 4 5 int[] A= {1,1,1,2,2,3}; 6 int n = A.length; 7 int index=removeDuplicates(A,n); 8 int[] B=new int[index]; //定义好数组长度不好改 9 for(int i=0;i<index;i++) { 10 B[i]=A[i]; 11 System.out.println(B[i]); 12 } 13 14 } 15 public static int removeDuplicates(int A[], int n) { 16 if (n == 0) return 0; 17 int occur = 1; 18 int index = 0; 19 for (int i = 1; i < n; i++) { 20 if (A[index] == A[i]) { 21 if (occur < 2) { 22 A[++index] = A[i]; 23 occur++; 24 } 25 } else { 26 A[++index] = A[i]; 27 occur = 1; 28 } 29 } 30 31 return index + 1; 32 } 33 }
Remove Duplicates from Sorted Array II
标签：follow dup for ++ UNC remove XA stat nbsp
原文地址：https://www.cnblogs.com/ncznx/p/9167081.html
