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.
Clarification:
Confused why the returned value is an integer but your answer is an array?
Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.
Internally you can think of this:
// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);
// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
print(nums[i]);
}
法一:暴力清除容器
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
set<int> res;
for(int num : nums){
res.insert(num);
}
vector <int>().swap(nums); //清空容器
for(int num : res){
nums.push_back(num);
}
return res.size();
}
};
法二:简单,但速度不是最快的
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int count = 0;
int sz = nums.size();
for(int i = 1; i < sz; i++){
if(nums[i - 1] == nums[i])
count++;
else
nums[i - count] = nums[i];
}
return sz - count;
}
};
法三:打败99%
class Solution {
public:
int removeDuplicates(vector<int>& nums) {
int cnt = 0;
int tmp = INT_MAX;
for (auto i = 0; i < nums.size(); i++){
if (tmp != nums[i]){
tmp = nums[cnt] = nums[i];
cnt++;
}
}
return cnt;
}
};

本文介绍了一种在不使用额外空间的情况下去除已排序数组中重复元素的方法,并提供了三种不同的C++实现方案,包括暴力清除容器法、简单高效法及最优实践法。
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