26. Remove Duplicates from Sorted Array

原题链接：https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/
这道题目是经典的已排序的数组去重问题。之前我有做过看过类似题目的，可以却忘记方法了，所以这次又是写了一个简单粗暴的方法：即遇到重复的元素，对数组进行移位操作：

import java.util.Arrays;

/**
 * Created by clearbug on 2018/2/26.
 */
public class Solution {

    public static void main(String[] args) {
        Solution s = new Solution();

        int[] arr = new int[]{1, 1, 2};
        System.out.println(s.removeDuplicates(arr));
        System.out.println(Arrays.toString(arr));

        arr = new int[]{1, 1, 2, 3, 3, 5, 6, 8, 8};
        System.out.println(s.removeDuplicates(arr));
        System.out.println(Arrays.toString(arr));

        arr = new int[]{1, 1, 1};
        System.out.println(s.removeDuplicates(arr));
        System.out.println(Arrays.toString(arr));
    }

    public int removeDuplicates(int[] nums) {
        if (nums.length < 2) { // nums.length is 0 or 1.
            return nums.length;
        }

        int len = nums.length;

        int mark = 0;
        int temp = nums[mark];
        for (int i = mark + 1; i < len;) {
            if (nums[i] == temp) {
                for (int j = i + 1; j < len; j++) {
                    nums[mark + (j - i)] = nums[j];
                }
                len--;
            } else {
                mark = i;
                temp = nums[mark];
                i++;
            }
        }

        return len;
    }

}

看了下提交结果：66 ms，beats 4.25%
这个结果好尴尬啊，然后我又去看了下官方提供的答案：

public int removeDuplicates(int[] nums) {
    if (nums.length == 0) return 0;
    int i = 0;
    for (int j = 1; j < nums.length; j++) {
        if (nums[j] != nums[i]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

看完之后只想说：尼玛，官方答案就是牛逼啊！