题目:
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.
思路:
顺序复制问题:如果当前元素大于目前结果中的最后一个元素,则添加该元素到当前结果中,并更新长度;否则直接跳过。时间复杂度O(n)。
代码:
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
if (nums.size() == 0)
return 0;
int new_length = 1;
int last_value = nums[0];
for (size_t i = 1; i < nums.size(); i++)
{
if (nums[i] > last_value) // the new element should be added
{
nums[new_length++] = nums[i];
last_value = nums[i];
}
}
return new_length;
}
};
代码: