Leetcode 26 Remove Duplicates from Sorted Array

最新推荐文章于 2024-10-06 13:06:38 发布

wanttifa

最新推荐文章于 2024-10-06 13:06:38 发布

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本文介绍了一种在不使用额外空间的情况下，去除已排序数组中重复元素的方法。通过一个计数器index，遍历数组，当遇到不同的元素时，将其放置在nums[index]的位置，最终返回index的大小加一作为新长度。

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Description

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]);
}

解题思路

这个题意很简单，就是求出数组中有几个不相同的元素。唯一的要求就是不能新建一个数组，要保证O(1)的空间复杂度。

既然不能新建数组，那么就设置一个计数器index。

然后遍历一遍数组，遍历过程中，比较nums[index]和nums[i]是否相同，若不相同，就把新的元素放在nums[index]位置上，最后输出index的大小就好了。

代码

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
    	//判断是否为空数组
        if(nums.empty()) return 0;
        int index=0;
        for(int i=1;i<nums.size();i++)
        {
            if(nums[index]!=nums[i])
                nums[++index]=nums[i];
        }
        return index+1;
    }
};