My code:

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        int del = 0;
        int count = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1]) {
                count++;
                if (count > 2)
                    del++;
                else
                    nums[i - del] = nums[i];
            }
            else {
                count = 1;
                nums[i - del] = nums[i];
            }
        }
        return nums.length - del;
    }
    
    public static void main(String[] args) {
        Solution test = new Solution();
        int[] a = {1, 1, 1, 1, 3, 3};
        System.out.println(test.removeDuplicates(a));
    }
}

My test result:

這次作業(yè)不難井辜，主要是直接看了此題的簡(jiǎn)單版本的代碼绎谦，以前做過(guò)。然后有了思路粥脚。這題變形在窃肠，允許最多出現(xiàn)兩次，那么就用一個(gè)count來(lái)計(jì)數(shù)就行了阿逃，超過(guò)兩次就開始刪除铭拧。然后不超過(guò)又相等的話赃蛛，這里就有個(gè)細(xì)節(jié)恃锉，就需要不停地往前移搀菩，往前覆蓋。

**
總結(jié)： Array, two pointer
**

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        int del = 0;
        int counter = 1; // used to record the number of elements with same values
        /** traverse the whole array */
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1]) {
                if (counter >= 2) {
                    del++; // if count >= 2, increase del by 1 
                }
                else {
                    nums[i - del] = nums[i]; // if not, move repetive element to front
                }
                counter++; // count the repetive elements
            }
            else {
                nums[i - del] = nums[i];
                counter = 1; // resotre counter
            }
        }
        return nums.length - del;
    }
}

差不多的做法破托，沒什么好說(shuō)的肪跋。

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int del = 0;
        int counter = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] == nums[i - 1]) {
                counter++;
                if (counter > 2) {
                    del++;
                }
                else {
                    nums[i - del] = nums[i];
                }
            }
            else {
                counter = 1;
                nums[i - del] = nums[i];
            }
        }
        
        return nums.length - del;
    }
}

差不多的解法。

Anyway, Good luck, Richardo! -- 09/04/2016