[Week 3] [Leetcode] [Array] Remove Duplicates from Sorted Array

GeeGeeZi

于 2017-06-16 17:23:53 发布

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本文链接：https://blog.youkuaiyun.com/GeeGeeZi/article/details/73342463

本文介绍了一种在不使用额外空间的情况下从已排序数组中删除重复元素的方法。通过一个简单的C++示例展示了如何利用索引遍历数组并保持元素唯一性，最终返回新数组的有效长度。

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Question:
Given a sorted array, remove the duplicates in place such that each element appear onlyonce 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 being1 and 2 respectively. It doesn't matter what you leave beyond the new length.
Analysis:

目标：根据题目了解到需要去除数组中重复的元素，并且整个运算过程不重新分配内存。

条件分析：首先这是一个已经排好序的数组，所以首先判断数组是否为空，如为空则直接返回0.其次，当数组不为空时，

Code:

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        //考虑数组为空数组的情况，返回0
        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;
    }
};