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.
给定一个排序数组,你需要在 原地 删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度。
不要使用额外的数组空间,你必须在 原地 修改输入数组 并在使用 O(1) 额外空间的条件下完成。
给定的是有序数组,如果数组中存在重复元素,那么一定是连续出现的。所以,我们这里使用两个指针:一个指向第一个元素idx,表明不重复子数组的右边界;另一个从第二个元素开始进行遍历,i。
完整代码:
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.empty()) return 0;
int idx = 0;
for (int i=1; i< nums.size(); i++){
if (nums[i] != nums[idx])
nums[++idx] = nums[i];
}
return idx+1;
}
};