LeetCode題目鏈接

題目：

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.

答案一（利用容器）：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int ns = nums.size();
        if (ns == 0) {
            return 0;
        }
        int i = 0;
        while (i < ns - 1) {
            if (nums[i] == nums[i + 1]) {
                nums.erase(nums.begin() + i + 1);
                ns--;
            } else {
                i++;
            }
        }
        
        return ns;
    }
};

答案二（Java)：

雙指針。直接改變數組，當出現重復的數的時候，跳過；當數不重復時，賦值給較慢的指針，最終的結果是數組前面一部分沒有重復的值，后面的數字無意義，由于會返回數組的大小。因此還是可以區分。

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