Question

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.
Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

**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 4 respectively.
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, which means a modification to the input array will be known to the caller as well.
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]);
}

Answer

**双指针法: ** 首先这是有序的数组，那么如果出现重复，就一定是重复数字的下一个(只会重复一次)。那么我就只需要比较当前位和下一位是否产生重复，如果产生了重复。那么我就用下一位来覆盖当前位，最后返回的数组长度。时间复杂度为O(n),空间复杂度为O(1),具体代码如下:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.size() == 0) return 0;
        int i =0 ,j = 1;
        while (j < nums.size()) {
            if (nums[i] != nums[j])
                nums[++i] = nums[j];
            j++;
        }
        return i+1;
    }
};

Rabbet

删除有序数组的重复项

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