【LeetCode】26. Remove Duplicates from Sorted Array

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本文介绍了一种在原地修改输入数组以去除重复元素的方法，并提供两种解决方案：一种使用内置函数unique，另一种通过双指针法手动实现。双指针法在不使用额外空间的情况下达到了O(n)的时间复杂度。

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题目：

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 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 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]);
}

描述：

给出一个有序数组，去除所有重复元素

分析：

第一步想到的是unique函数...

当然，用双指针法也能实现，不过感觉题意描述的貌似有点模糊

反正就是只要最前面是不重复的元素就行，系统自动有输出控制系统..

贡献两组数据：

[]
[1]

最精简版本代码：（不知道函数的内部实现，但是效率不是太高）

class Solution {
public:
	int removeDuplicates(vector<int>& nums) {
		return unique(nums.begin(), nums.end()) - nums.begin();
	}
};

手动实现：（时间复杂度O（n））

class Solution {
public:
	int removeDuplicates(vector<int>& nums) {
		if (nums.size() < 2) {
			return nums.size();
		}
		
        int new_index = 0;
		for (int i = 1; i < nums.size(); ++ i) {
			if (nums[i] != nums[new_index]) {
				++ new_index;
				nums[new_index] = nums[i];
			}
		}
		return new_index + 1;
	}
};