26. Remove Duplicates from Sorted Array

Easy

Given a sorted array nums, 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 by modifying the input array 

Example 1:

Given 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 returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5
 
, with the first five elements of nums being modified to 0, 1, 2, 3, and 4respectively. It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

解法：從有序陣列中去除重複項，使用快慢指標來記錄遍歷的座標，兩個指標都指向第一個數字，如果兩個指標指的數字相同，則快指標向前走一步，如果不同，則兩個指標都向前走一步，當快指標走完整個陣列後，慢指標當前的座標加1就是陣列中不同數字的個數，程式碼如下：

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.empty()) return 0;
        int pre = 0, cur = 0, n = nums.size();
        while (cur < n) {
            if (nums[pre] == nums[cur]) ++cur;
            else nums[++pre] = nums[cur++];
        }
        return pre + 1;
    }
};

執行結果：