lintcode-easy-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].

 

按理说不太难，但是做过的题有时候还是一下写不对……

分情况讨论：

当重复个数只有1个的时候，如果j对应的数是重复的，重复个数count加1

当重复个数为2的时候，找到下一个不重复的数，加到末尾

public class Solution {
    /**
     * @param A: a array of integers
     * @return : return an integer
     */
    public int removeDuplicates(int[] nums) {
        // write your code here
        if(nums == null)
            return 0;
        if(nums.length <= 2)
            return nums.length;
        
        int i = 1;
        int j = 1; 
        int count = 1;
        
        while(j < nums.length){
            if(count == 1){
                if(nums[j] == nums[j - 1]){
                    count++;
                    nums[i] = nums[j];
                    i++;
                    j++;
                }
                else{
                    nums[i] = nums[j];
                    i++;
                    j++;
                }
            }
            else{
                while(j < nums.length && nums[j] == nums[j - 1])
                    j++;
                if(j < nums.length){
                    nums[i] = nums[j];
                    count = 1;
                    i++;
                    j++;
                }
            }
        }
        
        return i;
    }
}