LeetCode 26 - Remove Duplicates from Sorted Array
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一、问题描述
Description: Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
For example: Given input array 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 new length.
Note: Do not allocate extra space for another array, you must do this in place with constant memory.
给一个有序数组,原地移除重复的元素,并返回新数组的长度。
注意:只能原地工作,不能使用额外的数组。
二、解题报告
由于输入的是有序数组,在遍历过程中,我们只需要判断是否与前面的重复即可。
代码:
class Solution {public: int removeDuplicates(vector<int>& nums) { if(nums.empty()) return 0; int count = 1; int pre = nums[0]; vector<int>::iterator beg = nums.begin(); for(++beg; beg!=nums.end(); ) { // 从第二个开始 if(*beg == pre) // 重复,移除 beg = nums.erase(beg); else // 不重复,更新pre { pre = *beg; ++beg; } } return nums.size(); }};
LeetCode答案源代码:https://github.com/SongLee24/LeetCode
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