/**
* Source : https://oj.leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
*
*
* 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].
*
*/
public class RemoveDuplicates2 {
/**
* 只需要找到不重复的元素个数,不重复的定义是:小于等于2个
* 遍历数组
*
* @param arr
* @return
*/
public int remove (int[] arr) {
if (arr.length <= 2) {
return arr.length;
}
int count = 1;
int pos = 0;
for (int i = 1; i < arr.length; i++) {
if (arr[i] == arr[i-1]) {
count ++;
} else {
if (count > 2) {
pos += 2;
} else {
pos += count;
}
count = 1;
}
}
if (count > 2) {
pos += 2;
} else {
pos += count;
}
return pos;
}
/**
* 出现3次及以上才算重复
* 第一次出现记一次数。第二次出现记一次数
*
* @param arr
* @return
*/
public int remove1 (int[] arr) {
if (arr.length <= 2) {
return arr.length;
}
int count = 1;
int pos = 0;
for (int i = 1; i < arr.length; i++) {
if (arr[i] == arr[i-1]) {
count ++;
if (count == 2) {
pos ++;
}
} else {
count = 1;
pos ++;
}
}
return pos +1;
}
public static void main(String[] args) {
RemoveDuplicates2 removeDuplicates = new RemoveDuplicates2();
int[] arr0 = new int[]{1,1,1};
int[] arr = new int[]{1,1,2};
int[] arr1 = new int[]{1,1,1,2};
int[] arr2 = new int[]{1,1,22,22,22,33};
int[] arr3 = new int[]{1,1,1,1,1,1};
System.out.println(removeDuplicates.remove(arr0) + "----" + removeDuplicates.remove1(arr0));
System.out.println(removeDuplicates.remove(arr) + "----" + removeDuplicates.remove1(arr));
System.out.println(removeDuplicates.remove(arr1) + "----" + removeDuplicates.remove1(arr1));
System.out.println(removeDuplicates.remove(arr2) + "----" + removeDuplicates.remove1(arr2));
System.out.println(removeDuplicates.remove(arr3) + "----" + removeDuplicates.remove1(arr3));
}
}