【Leetcode】Remove Duplicates from Sorted Array
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描述:
Given a sorted array, 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 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 being 1
and 2
respectively. It doesn't matter what you leave beyond the new length.
解法1:
给两个索引,
oldIdx 用来从第二个元素开始遍历原数组,
newIdx 保存已经过滤重复元素数组的最后一个元素的下标。
在循环内部,比较 数组在oIdx 和 nIdx 的值是否相同,如果相同,oIdx++, 如果不相同, 先让nIdx++,表示无重复元素数组需要添加一个元素,然后将arr[ oIdx] 赋给 arr[nIdx] 即可,再将 oIdx++。直到遍历完原数组。
class Solution {public: int removeDuplicates(vector<int>& nums) { if( nums.size() == 0) return 0; int newIdx = 0; int oldIdx = 1; while(oldIdx < nums.size()) { if(nums[newIdx] != nums[oldIdx]) { newIdx++; nums[newIdx] = nums[oldIdx]; } oldIdx++; } nums.resize(oldIdx+1); return newIdx+1; }};
解法2:
利用STL库函数distance() 求出两个迭代器之间的距离, 和 unique() 函数 去掉一个容器中的重复的连续元素。
class Solution {public: int removeDuplicates(vector<int>& nums) { if(nums.size() == 0) return 0; return distance(nums.begin(), unique(nums.begin(), nums.begin() + nums.size())); }};
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